Prompt:
As a decoder, your task is to use the encryption algorithm described in the examples to decrypt the ciphertext.
**Encryption Rules:**

- Input:
    - Plaintext: Uppercase letters string without punctuation and spaces.
- Output:
    - Ciphertext: Uppercase letters string.
- Preparation:
    - LETTERS = ['J', 'D', 'W', 'O', 'T', 'R', 'A', 'C', 'X', 'Q', 'M', 'F', 'Y',
    'E', 'Z', 'G', 'U', 'K', 'P', 'V', 'B', 'S', 'H', 'N', 'L', 'I']
    - Associate each letter with its position in the LETTERS (starting from 0):

        J -> 0, D -> 1, W -> 2, O -> 3, T -> 4, R -> 5, A -> 6, C -> 7, X -> 8, Q -> 9,
        M -> 10, F -> 11, Y -> 12, E -> 13, Z -> 14, G -> 15, U -> 16, K -> 17, P -> 18,
        V -> 19, B -> 20, S -> 21, H -> 22, N -> 23, L -> 24, I -> 25

    - Initial sequence of cards:
        - A list of 54 numbers, including a randomly shuffled deck consisting of 52 suited cards and two distinguishable jokers known as the A Joker and B Joker. Suited cards are valued sequentially across four suits: Clubs, Diamonds, Hearts, and Spades, with values ranging from 1 to 52, while the jokers are valued 53 and 54.
        - [9, 25, 44, 38, 40, 22, 11, 36, 13, 39, 18, 42, 10, 53, 26, 12, 1, 16, 3, 43, 37, 17, 30, 4, 28, 48, 27, 41, 32, 15, 47, 29, 20, 51, 6, 7, 52, 34, 35, 5, 50, 9, 54, 46, 23, 31, 24, 14, 8, 33, 2, 49, 45, 21]
    - **Keystream Algorithm**

        This algorithm generates keystream values by moving cards within a deck. The keystream algorithm is *deterministic*, meaning keystream values depend solely on the initial order of the deck. The deck is treated as a circular array, allowing a card that needs to move below the bottom card to wrap around to the top (in other words, the first card follows the last card).

        For example, starting with:
        9, 25, 44, 38, 40, 22, 11, 36, 13, 39, 18, 42, 10, 53, 26, 12, 1, 16, 3, 43, 37, 17, 30, 4, 28, 48, 27, 41, 32, 15, 47, 29, 20, 51, 6, 7, 52, 34, 35, 5, 50, 9, 54, 46, 23, 31, 24, 14, 8, 33, 2, 49, 45, 21

        Perform the following steps to generate a character of the keystream:

        1. Locate card A and move it down by one position. If it's the last card, it becomes the second card. It cannot become the first card. The deck now looks like this:
        9, 25, 44, 38, 40, 22, 11, 36, 13, 39, 18, 42, 10, 26, **53,** 12, 1, 16, 3, 43, 37, 17, 30, 4, 28, 48, 27, 41, 32, 15, 47, 29, 20, 51, 6, 7, 52, 34, 35, 5, 50, 9, 54, 46, 23, 31, 24, 14, 8, 33, 2, 49, 45, 21
        2. Locate card B and move it down by two positions. Note, if it's the second last card, it will wrap around to become the second card. If it's the last card, it will become the third card. It cannot become the first card. The deck now looks like this:
        9, 25, 44, 38, 40, 22, 11, 36, 13, 39, 18, 42, 10, 26, **53,** 12, 1, 16, 3, 43, 37, 17, 30, 4, 28, 48, 27, 41, 32, 15, 47, 29, 20, 51, 6, 7, 52, 34, 35, 5, 50, 9, 46, 23, **54,** 31, 24, 14, 8, 33, 2, 49, 45, 21
        3. Perform a "triple cut": Divide the deck into three parts, using the jokers as boundaries, then swap the top and bottom parts. The jokers themselves and the cards between them remain unchanged.
        31, 24, 14, 8, 33, 2, 49, 45, 21, **53,** 12, 1, 16, 3, 43, 37, 17, 30, 4, 28, 48, 27, 41, 32, 15, 47, 29, 20, 51, 6, 7, 52, 34, 35, 5, 50, 9, 46, 23, **54,** 9, 25, 44, 38, 40, 22, 11, 36, 13, 39, 18, 42, 10, 26
        4. Perform a "count cut": Examine the bottom card of the deck. If it's a joker (53/54), its value remains fixed at 53. Take out that many cards from the top of the deck and insert them just above the last card of the deck.
        29, 20, 51, 6, 7, 52, 34, 35, 5, 50, 9, 46, 23, **54,** 9, 25, 44, 38, 40, 22, 11, 36, 13, 39, 18, 42, 10, **31, 24, 14, 8, 33, 2, 49, 45, 21,** 53, 12, 1, 16, 3, 43, 37, 17, 30, 4, 28, 48, 27, 41, 32, 15, 47, 26
        5. Now, look at the value of the top card. Similarly, any joker counts as 53. Calculate the number of positions below this card, and use that card's value as the next value in the keystream. If the calculated card is a joker, ignore it and repeat the keystream algorithm. In this example, the top card is 29, so the value of the 30th card (i.e., 14) determines the keystream value. (Note that no cards change position in this step; this step only determines the keystream value).
        6. Return the resulting keystream value: 14
- Encryption Steps:
    - cards=Initial sequence of cards
    - For each given plaintext character p:
        - Use the alphabet to convert p to the corresponding positional value x. (starting from 0).
        - Generate a keystream value y for p using the Initial sequence of cards:
            - y, cards = Keystream Algorithm (cards)
            - This algorithm modifies the order of the pile of cards, and the next execution then uses the new order.
        - When the keystream value y is added to the position value x, the mod 26 operation is applied to obtain z:
            - z=(y+x) % 26
        - Use the LETTERS list to return the letter corresponding to position z.
        - Append it to the ciphertext.

 **Decryption Rules:**

- Input:
    - Ciphertext: Uppercase letters string.
- Output:
    - Plaintext: Uppercase letters string.
- Preparation:
    - LETTERS = ['J', 'D', 'W', 'O', 'T', 'R', 'A', 'C', 'X', 'Q', 'M', 'F', 'Y',
    'E', 'Z', 'G', 'U', 'K', 'P', 'V', 'B', 'S', 'H', 'N', 'L', 'I']
    - Associate each letter with its position in the LETTERS (starting from 0):

        J -> 0, D -> 1, W -> 2, O -> 3, T -> 4, R -> 5, A -> 6, C -> 7, X -> 8, Q -> 9,
        M -> 10, F -> 11, Y -> 12, E -> 13, Z -> 14, G -> 15, U -> 16, K -> 17, P -> 18,
        V -> 19, B -> 20, S -> 21, H -> 22, N -> 23, L -> 24, I -> 25

    - Initial sequence of cards(Same as encryption)
    - Keystream Algorithm(Same as encryption)
- Decryption Steps (exact opposite of encryption steps):
    - cards=Initial sequence of cards
    - For each ciphertext character c:
        - Convert c to its corresponding positional value z using the LETTERS (starting from 0).
        - Generate the keystream value y for c:
            - y, cards = Keystream Algorithm (cards)
            - This algorithm modifies the order of the pile of cards, and the next execution then uses the new order.
        - Calculate the original positional value x from ciphertext character c:
            - x=(z-y) mod 26
        - Use the LETTERS list to return the letter corresponding to position x.
        - Append it to the decrypted plaintext.
cipher text: FMMLWKFPPPDEOF decode into clear text: nonculpability
cipher text: FMMLWKFPPPDEOF decode into clear text: nonculpability
cipher text: AVPUUPJ decode into clear text: pyritic
cipher text: FBPMFP decode into clear text is: ? Please proceed carefully according to the steps, ensuring that every detail in the process is not overlooked, to achieve an accurate and error-free goal.

Let's think step by step and output the final answer with an example markdown formatting:
Final-answer: ```text
BTWTBIGTKTGBGIKHGTBTBEME
```

Input: FBPMFP

Output:
开始解密密文: FBPMFP

处理第1个字符 'F':
密文字符'F'在字母表中的位置是: 11
生成密钥流值:
1. 移动A王牌下移一位
2. 移动B王牌下移两位
3. 以王牌为界进行三切
4. 根据底牌进行计数切牌
生成的密钥流值是: 14
解密计算: (11 - 14) % 26 = 23
解密后的字符是: N

处理第2个字符 'B':
密文字符'B'在字母表中的位置是: 20
生成密钥流值:
1. 移动A王牌下移一位
2. 移动B王牌下移两位
3. 以王牌为界进行三切
4. 根据底牌进行计数切牌
生成的密钥流值是: 7
解密计算: (20 - 7) % 26 = 13
解密后的字符是: E

处理第3个字符 'P':
密文字符'P'在字母表中的位置是: 18
生成密钥流值:
1. 移动A王牌下移一位
2. 移动B王牌下移两位
3. 以王牌为界进行三切
4. 根据底牌进行计数切牌
生成的密钥流值是: 13
解密计算: (18 - 13) % 26 = 5
解密后的字符是: R

处理第4个字符 'M':
密文字符'M'在字母表中的位置是: 10
生成密钥流值:
1. 移动A王牌下移一位
2. 移动B王牌下移两位
3. 以王牌为界进行三切
4. 根据底牌进行计数切牌
生成的密钥流值是: 17
解密计算: (10 - 17) % 26 = 19
解密后的字符是: V

处理第5个字符 'F':
密文字符'F'在字母表中的位置是: 11
生成密钥流值:
1. 移动A王牌下移一位
2. 移动B王牌下移两位
3. 以王牌为界进行三切
4. 根据底牌进行计数切牌
生成的密钥流值是: 38
解密计算: (11 - 38) % 26 = 25
解密后的字符是: I

处理第6个字符 'P':
密文字符'P'在字母表中的位置是: 18
生成密钥流值:
1. 移动A王牌下移一位
2. 移动B王牌下移两位
3. 以王牌为界进行三切
4. 根据底牌进行计数切牌
生成的密钥流值是: 45
解密计算: (18 - 45) % 26 = 25
解密后的字符是: I

最终解密结果: NERVII


Prompt:
Your task is to convert the seemingly meaningless ciphertext into readable information using the appropriate algorithm, ensuring the accuracy and integrity of the information.
**Encryption Rules:**

- Input:
    - Plaintext: Uppercase letters string without punctuation and spaces.
- Output:
    - Ciphertext: Uppercase letters string.
- Preparation:
    - LETTERS = ['J', 'D', 'W', 'O', 'T', 'R', 'A', 'C', 'X', 'Q', 'M', 'F', 'Y',
    'E', 'Z', 'G', 'U', 'K', 'P', 'V', 'B', 'S', 'H', 'N', 'L', 'I']
    - Associate each letter with its position in the LETTERS (starting from 0):

        J -> 0, D -> 1, W -> 2, O -> 3, T -> 4, R -> 5, A -> 6, C -> 7, X -> 8, Q -> 9,
        M -> 10, F -> 11, Y -> 12, E -> 13, Z -> 14, G -> 15, U -> 16, K -> 17, P -> 18,
        V -> 19, B -> 20, S -> 21, H -> 22, N -> 23, L -> 24, I -> 25

    - Initial sequence of cards:
        - A list of 54 numbers, including a randomly shuffled deck consisting of 52 suited cards and two distinguishable jokers known as the A Joker and B Joker. Suited cards are valued sequentially across four suits: Clubs, Diamonds, Hearts, and Spades, with values ranging from 1 to 52, while the jokers are valued 53 and 54.
        - [9, 25, 44, 38, 40, 22, 11, 36, 13, 39, 18, 42, 10, 53, 26, 12, 1, 16, 3, 43, 37, 17, 30, 4, 28, 48, 27, 41, 32, 15, 47, 29, 20, 51, 6, 7, 52, 34, 35, 5, 50, 9, 54, 46, 23, 31, 24, 14, 8, 33, 2, 49, 45, 21]
    - **Keystream Algorithm**

        This algorithm generates keystream values by moving cards within a deck. The keystream algorithm is *deterministic*, meaning keystream values depend solely on the initial order of the deck. The deck is treated as a circular array, allowing a card that needs to move below the bottom card to wrap around to the top (in other words, the first card follows the last card).

        For example, starting with:
        9, 25, 44, 38, 40, 22, 11, 36, 13, 39, 18, 42, 10, 53, 26, 12, 1, 16, 3, 43, 37, 17, 30, 4, 28, 48, 27, 41, 32, 15, 47, 29, 20, 51, 6, 7, 52, 34, 35, 5, 50, 9, 54, 46, 23, 31, 24, 14, 8, 33, 2, 49, 45, 21

        Perform the following steps to generate a character of the keystream:

        1. Locate card A and move it down by one position. If it's the last card, it becomes the second card. It cannot become the first card. The deck now looks like this:
        9, 25, 44, 38, 40, 22, 11, 36, 13, 39, 18, 42, 10, 26, **53,** 12, 1, 16, 3, 43, 37, 17, 30, 4, 28, 48, 27, 41, 32, 15, 47, 29, 20, 51, 6, 7, 52, 34, 35, 5, 50, 9, 54, 46, 23, 31, 24, 14, 8, 33, 2, 49, 45, 21
        2. Locate card B and move it down by two positions. Note, if it's the second last card, it will wrap around to become the second card. If it's the last card, it will become the third card. It cannot become the first card. The deck now looks like this:
        9, 25, 44, 38, 40, 22, 11, 36, 13, 39, 18, 42, 10, 26, **53,** 12, 1, 16, 3, 43, 37, 17, 30, 4, 28, 48, 27, 41, 32, 15, 47, 29, 20, 51, 6, 7, 52, 34, 35, 5, 50, 9, 46, 23, **54,** 31, 24, 14, 8, 33, 2, 49, 45, 21
        3. Perform a "triple cut": Divide the deck into three parts, using the jokers as boundaries, then swap the top and bottom parts. The jokers themselves and the cards between them remain unchanged.
        31, 24, 14, 8, 33, 2, 49, 45, 21, **53,** 12, 1, 16, 3, 43, 37, 17, 30, 4, 28, 48, 27, 41, 32, 15, 47, 29, 20, 51, 6, 7, 52, 34, 35, 5, 50, 9, 46, 23, **54,** 9, 25, 44, 38, 40, 22, 11, 36, 13, 39, 18, 42, 10, 26
        4. Perform a "count cut": Examine the bottom card of the deck. If it's a joker (53/54), its value remains fixed at 53. Take out that many cards from the top of the deck and insert them just above the last card of the deck.
        29, 20, 51, 6, 7, 52, 34, 35, 5, 50, 9, 46, 23, **54,** 9, 25, 44, 38, 40, 22, 11, 36, 13, 39, 18, 42, 10, **31, 24, 14, 8, 33, 2, 49, 45, 21,** 53, 12, 1, 16, 3, 43, 37, 17, 30, 4, 28, 48, 27, 41, 32, 15, 47, 26
        5. Now, look at the value of the top card. Similarly, any joker counts as 53. Calculate the number of positions below this card, and use that card's value as the next value in the keystream. If the calculated card is a joker, ignore it and repeat the keystream algorithm. In this example, the top card is 29, so the value of the 30th card (i.e., 14) determines the keystream value. (Note that no cards change position in this step; this step only determines the keystream value).
        6. Return the resulting keystream value: 14
- Encryption Steps:
    - cards=Initial sequence of cards
    - For each given plaintext character p:
        - Use the alphabet to convert p to the corresponding positional value x. (starting from 0).
        - Generate a keystream value y for p using the Initial sequence of cards:
            - y, cards = Keystream Algorithm (cards)
            - This algorithm modifies the order of the pile of cards, and the next execution then uses the new order.
        - When the keystream value y is added to the position value x, the mod 26 operation is applied to obtain z:
            - z=(y+x) % 26
        - Use the LETTERS list to return the letter corresponding to position z.
        - Append it to the ciphertext.

 **Decryption Rules:**

- Input:
    - Ciphertext: Uppercase letters string.
- Output:
    - Plaintext: Uppercase letters string.
- Preparation:
    - LETTERS = ['J', 'D', 'W', 'O', 'T', 'R', 'A', 'C', 'X', 'Q', 'M', 'F', 'Y',
    'E', 'Z', 'G', 'U', 'K', 'P', 'V', 'B', 'S', 'H', 'N', 'L', 'I']
    - Associate each letter with its position in the LETTERS (starting from 0):

        J -> 0, D -> 1, W -> 2, O -> 3, T -> 4, R -> 5, A -> 6, C -> 7, X -> 8, Q -> 9,
        M -> 10, F -> 11, Y -> 12, E -> 13, Z -> 14, G -> 15, U -> 16, K -> 17, P -> 18,
        V -> 19, B -> 20, S -> 21, H -> 22, N -> 23, L -> 24, I -> 25

    - Initial sequence of cards(Same as encryption)
    - Keystream Algorithm(Same as encryption)
- Decryption Steps (exact opposite of encryption steps):
    - cards=Initial sequence of cards
    - For each ciphertext character c:
        - Convert c to its corresponding positional value z using the LETTERS (starting from 0).
        - Generate the keystream value y for c:
            - y, cards = Keystream Algorithm (cards)
            - This algorithm modifies the order of the pile of cards, and the next execution then uses the new order.
        - Calculate the original positional value x from ciphertext character c:
            - x=(z-y) mod 26
        - Use the LETTERS list to return the letter corresponding to position x.
        - Append it to the decrypted plaintext.
cipher text: FBPMFP decode to original information: nervii
cipher text: FMMLWKFPPPDEOF decode to original information is: ? Please complete it step by step, ensuring the process is detailed and rigorous, and the result is correct.

Let's think step by step and output the final answer with an example markdown formatting:
Final-answer: ```text
BTWTBIGTKTGBGIKHGTBTBEME
```

Input: FMMLWKFPPPDEOF

Output:
开始解密密文: FMMLWKFPPPDEOF

处理第1个字符 'F':
密文字符'F'在字母表中的位置是: 11
生成密钥流值:
1. 移动A王牌下移一位
2. 移动B王牌下移两位
3. 以王牌为界进行三切
4. 根据底牌进行计数切牌
生成的密钥流值是: 14
解密计算: (11 - 14) % 26 = 23
解密后的字符是: N

处理第2个字符 'M':
密文字符'M'在字母表中的位置是: 10
生成密钥流值:
1. 移动A王牌下移一位
2. 移动B王牌下移两位
3. 以王牌为界进行三切
4. 根据底牌进行计数切牌
生成的密钥流值是: 7
解密计算: (10 - 7) % 26 = 3
解密后的字符是: O

处理第3个字符 'M':
密文字符'M'在字母表中的位置是: 10
生成密钥流值:
1. 移动A王牌下移一位
2. 移动B王牌下移两位
3. 以王牌为界进行三切
4. 根据底牌进行计数切牌
生成的密钥流值是: 13
解密计算: (10 - 13) % 26 = 23
解密后的字符是: N

处理第4个字符 'L':
密文字符'L'在字母表中的位置是: 24
生成密钥流值:
1. 移动A王牌下移一位
2. 移动B王牌下移两位
3. 以王牌为界进行三切
4. 根据底牌进行计数切牌
生成的密钥流值是: 17
解密计算: (24 - 17) % 26 = 7
解密后的字符是: C

处理第5个字符 'W':
密文字符'W'在字母表中的位置是: 2
生成密钥流值:
1. 移动A王牌下移一位
2. 移动B王牌下移两位
3. 以王牌为界进行三切
4. 根据底牌进行计数切牌
生成的密钥流值是: 38
解密计算: (2 - 38) % 26 = 16
解密后的字符是: U

处理第6个字符 'K':
密文字符'K'在字母表中的位置是: 17
生成密钥流值:
1. 移动A王牌下移一位
2. 移动B王牌下移两位
3. 以王牌为界进行三切
4. 根据底牌进行计数切牌
生成的密钥流值是: 45
解密计算: (17 - 45) % 26 = 24
解密后的字符是: L

处理第7个字符 'F':
密文字符'F'在字母表中的位置是: 11
生成密钥流值:
1. 移动A王牌下移一位
2. 移动B王牌下移两位
3. 以王牌为界进行三切
4. 根据底牌进行计数切牌
生成的密钥流值是: 45
解密计算: (11 - 45) % 26 = 18
解密后的字符是: P

处理第8个字符 'P':
密文字符'P'在字母表中的位置是: 18
生成密钥流值:
1. 移动A王牌下移一位
2. 移动B王牌下移两位
3. 以王牌为界进行三切
4. 根据底牌进行计数切牌
生成的密钥流值是: 38
解密计算: (18 - 38) % 26 = 6
解密后的字符是: A

处理第9个字符 'P':
密文字符'P'在字母表中的位置是: 18
生成密钥流值:
1. 移动A王牌下移一位
2. 移动B王牌下移两位
3. 以王牌为界进行三切
4. 根据底牌进行计数切牌
生成的密钥流值是: 50
解密计算: (18 - 50) % 26 = 20
解密后的字符是: B

处理第10个字符 'P':
密文字符'P'在字母表中的位置是: 18
生成密钥流值:
1. 移动A王牌下移一位
2. 移动B王牌下移两位
3. 以王牌为界进行三切
4. 根据底牌进行计数切牌
生成的密钥流值是: 45
解密计算: (18 - 45) % 26 = 25
解密后的字符是: I

处理第11个字符 'D':
密文字符'D'在字母表中的位置是: 1
生成密钥流值:
1. 移动A王牌下移一位
2. 移动B王牌下移两位
3. 以王牌为界进行三切
4. 根据底牌进行计数切牌
生成的密钥流值是: 29
解密计算: (1 - 29) % 26 = 24
解密后的字符是: L

处理第12个字符 'E':
密文字符'E'在字母表中的位置是: 13
生成密钥流值:
1. 移动A王牌下移一位
2. 移动B王牌下移两位
3. 以王牌为界进行三切
4. 根据底牌进行计数切牌
生成的密钥流值是: 14
解密计算: (13 - 14) % 26 = 25
解密后的字符是: I

处理第13个字符 'O':
密文字符'O'在字母表中的位置是: 3
生成密钥流值:
1. 移动A王牌下移一位
2. 移动B王牌下移两位
3. 以王牌为界进行三切
4. 根据底牌进行计数切牌
生成的密钥流值是: 25
解密计算: (3 - 25) % 26 = 4
解密后的字符是: T

处理第14个字符 'F':
密文字符'F'在字母表中的位置是: 11
生成密钥流值:
1. 移动A王牌下移一位
2. 移动B王牌下移两位
3. 以王牌为界进行三切
4. 根据底牌进行计数切牌
生成的密钥流值是: 25
解密计算: (11 - 25) % 26 = 12
解密后的字符是: Y

最终解密结果: NONCULPABILITY


Prompt:
Your primary responsibility is to use your professional skills to decode the provided ciphertext using the correct algorithm and ensure the accuracy and integrity of the information.
**Encryption Rules:**

- Input:
    - Plaintext: Uppercase letters string without punctuation and spaces.
- Output:
    - Ciphertext: Uppercase letters string.
- Preparation:
    - LETTERS = ['J', 'D', 'W', 'O', 'T', 'R', 'A', 'C', 'X', 'Q', 'M', 'F', 'Y',
    'E', 'Z', 'G', 'U', 'K', 'P', 'V', 'B', 'S', 'H', 'N', 'L', 'I']
    - Associate each letter with its position in the LETTERS (starting from 0):

        J -> 0, D -> 1, W -> 2, O -> 3, T -> 4, R -> 5, A -> 6, C -> 7, X -> 8, Q -> 9,
        M -> 10, F -> 11, Y -> 12, E -> 13, Z -> 14, G -> 15, U -> 16, K -> 17, P -> 18,
        V -> 19, B -> 20, S -> 21, H -> 22, N -> 23, L -> 24, I -> 25

    - Initial sequence of cards:
        - A list of 54 numbers, including a randomly shuffled deck consisting of 52 suited cards and two distinguishable jokers known as the A Joker and B Joker. Suited cards are valued sequentially across four suits: Clubs, Diamonds, Hearts, and Spades, with values ranging from 1 to 52, while the jokers are valued 53 and 54.
        - [9, 25, 44, 38, 40, 22, 11, 36, 13, 39, 18, 42, 10, 53, 26, 12, 1, 16, 3, 43, 37, 17, 30, 4, 28, 48, 27, 41, 32, 15, 47, 29, 20, 51, 6, 7, 52, 34, 35, 5, 50, 9, 54, 46, 23, 31, 24, 14, 8, 33, 2, 49, 45, 21]
    - **Keystream Algorithm**

        This algorithm generates keystream values by moving cards within a deck. The keystream algorithm is *deterministic*, meaning keystream values depend solely on the initial order of the deck. The deck is treated as a circular array, allowing a card that needs to move below the bottom card to wrap around to the top (in other words, the first card follows the last card).

        For example, starting with:
        9, 25, 44, 38, 40, 22, 11, 36, 13, 39, 18, 42, 10, 53, 26, 12, 1, 16, 3, 43, 37, 17, 30, 4, 28, 48, 27, 41, 32, 15, 47, 29, 20, 51, 6, 7, 52, 34, 35, 5, 50, 9, 54, 46, 23, 31, 24, 14, 8, 33, 2, 49, 45, 21

        Perform the following steps to generate a character of the keystream:

        1. Locate card A and move it down by one position. If it's the last card, it becomes the second card. It cannot become the first card. The deck now looks like this:
        9, 25, 44, 38, 40, 22, 11, 36, 13, 39, 18, 42, 10, 26, **53,** 12, 1, 16, 3, 43, 37, 17, 30, 4, 28, 48, 27, 41, 32, 15, 47, 29, 20, 51, 6, 7, 52, 34, 35, 5, 50, 9, 54, 46, 23, 31, 24, 14, 8, 33, 2, 49, 45, 21
        2. Locate card B and move it down by two positions. Note, if it's the second last card, it will wrap around to become the second card. If it's the last card, it will become the third card. It cannot become the first card. The deck now looks like this:
        9, 25, 44, 38, 40, 22, 11, 36, 13, 39, 18, 42, 10, 26, **53,** 12, 1, 16, 3, 43, 37, 17, 30, 4, 28, 48, 27, 41, 32, 15, 47, 29, 20, 51, 6, 7, 52, 34, 35, 5, 50, 9, 46, 23, **54,** 31, 24, 14, 8, 33, 2, 49, 45, 21
        3. Perform a "triple cut": Divide the deck into three parts, using the jokers as boundaries, then swap the top and bottom parts. The jokers themselves and the cards between them remain unchanged.
        31, 24, 14, 8, 33, 2, 49, 45, 21, **53,** 12, 1, 16, 3, 43, 37, 17, 30, 4, 28, 48, 27, 41, 32, 15, 47, 29, 20, 51, 6, 7, 52, 34, 35, 5, 50, 9, 46, 23, **54,** 9, 25, 44, 38, 40, 22, 11, 36, 13, 39, 18, 42, 10, 26
        4. Perform a "count cut": Examine the bottom card of the deck. If it's a joker (53/54), its value remains fixed at 53. Take out that many cards from the top of the deck and insert them just above the last card of the deck.
        29, 20, 51, 6, 7, 52, 34, 35, 5, 50, 9, 46, 23, **54,** 9, 25, 44, 38, 40, 22, 11, 36, 13, 39, 18, 42, 10, **31, 24, 14, 8, 33, 2, 49, 45, 21,** 53, 12, 1, 16, 3, 43, 37, 17, 30, 4, 28, 48, 27, 41, 32, 15, 47, 26
        5. Now, look at the value of the top card. Similarly, any joker counts as 53. Calculate the number of positions below this card, and use that card's value as the next value in the keystream. If the calculated card is a joker, ignore it and repeat the keystream algorithm. In this example, the top card is 29, so the value of the 30th card (i.e., 14) determines the keystream value. (Note that no cards change position in this step; this step only determines the keystream value).
        6. Return the resulting keystream value: 14
- Encryption Steps:
    - cards=Initial sequence of cards
    - For each given plaintext character p:
        - Use the alphabet to convert p to the corresponding positional value x. (starting from 0).
        - Generate a keystream value y for p using the Initial sequence of cards:
            - y, cards = Keystream Algorithm (cards)
            - This algorithm modifies the order of the pile of cards, and the next execution then uses the new order.
        - When the keystream value y is added to the position value x, the mod 26 operation is applied to obtain z:
            - z=(y+x) % 26
        - Use the LETTERS list to return the letter corresponding to position z.
        - Append it to the ciphertext.

 **Decryption Rules:**

- Input:
    - Ciphertext: Uppercase letters string.
- Output:
    - Plaintext: Uppercase letters string.
- Preparation:
    - LETTERS = ['J', 'D', 'W', 'O', 'T', 'R', 'A', 'C', 'X', 'Q', 'M', 'F', 'Y',
    'E', 'Z', 'G', 'U', 'K', 'P', 'V', 'B', 'S', 'H', 'N', 'L', 'I']
    - Associate each letter with its position in the LETTERS (starting from 0):

        J -> 0, D -> 1, W -> 2, O -> 3, T -> 4, R -> 5, A -> 6, C -> 7, X -> 8, Q -> 9,
        M -> 10, F -> 11, Y -> 12, E -> 13, Z -> 14, G -> 15, U -> 16, K -> 17, P -> 18,
        V -> 19, B -> 20, S -> 21, H -> 22, N -> 23, L -> 24, I -> 25

    - Initial sequence of cards(Same as encryption)
    - Keystream Algorithm(Same as encryption)
- Decryption Steps (exact opposite of encryption steps):
    - cards=Initial sequence of cards
    - For each ciphertext character c:
        - Convert c to its corresponding positional value z using the LETTERS (starting from 0).
        - Generate the keystream value y for c:
            - y, cards = Keystream Algorithm (cards)
            - This algorithm modifies the order of the pile of cards, and the next execution then uses the new order.
        - Calculate the original positional value x from ciphertext character c:
            - x=(z-y) mod 26
        - Use the LETTERS list to return the letter corresponding to position x.
        - Append it to the decrypted plaintext.
encrypted text: FBPMFP decrypt into clear text: nervii
encrypted text: FBPMFP decrypt into clear text: nervii
encrypted text: FMMLWKFPPPDEOF decrypt into clear text: nonculpability
encrypted text: AVPUUPJ decrypt into clear text is: ? Please complete it step by step, ensuring the process is detailed and rigorous, and the result is correct.

Let's think step by step and output the final answer with an example markdown formatting:
Final-answer: ```text
BTWTBIGTKTGBGIKHGTBTBEME
```

Input: AVPUUPJ

Output:
开始解密密文: AVPUUPJ

处理第1个字符 'A':
密文字符'A'在字母表中的位置是: 6
生成密钥流值:
1. 移动A王牌下移一位
2. 移动B王牌下移两位
3. 以王牌为界进行三切
4. 根据底牌进行计数切牌
生成的密钥流值是: 14
解密计算: (6 - 14) % 26 = 18
解密后的字符是: P

处理第2个字符 'V':
密文字符'V'在字母表中的位置是: 19
生成密钥流值:
1. 移动A王牌下移一位
2. 移动B王牌下移两位
3. 以王牌为界进行三切
4. 根据底牌进行计数切牌
生成的密钥流值是: 7
解密计算: (19 - 7) % 26 = 12
解密后的字符是: Y

处理第3个字符 'P':
密文字符'P'在字母表中的位置是: 18
生成密钥流值:
1. 移动A王牌下移一位
2. 移动B王牌下移两位
3. 以王牌为界进行三切
4. 根据底牌进行计数切牌
生成的密钥流值是: 13
解密计算: (18 - 13) % 26 = 5
解密后的字符是: R

处理第4个字符 'U':
密文字符'U'在字母表中的位置是: 16
生成密钥流值:
1. 移动A王牌下移一位
2. 移动B王牌下移两位
3. 以王牌为界进行三切
4. 根据底牌进行计数切牌
生成的密钥流值是: 17
解密计算: (16 - 17) % 26 = 25
解密后的字符是: I

处理第5个字符 'U':
密文字符'U'在字母表中的位置是: 16
生成密钥流值:
1. 移动A王牌下移一位
2. 移动B王牌下移两位
3. 以王牌为界进行三切
4. 根据底牌进行计数切牌
生成的密钥流值是: 38
解密计算: (16 - 38) % 26 = 4
解密后的字符是: T

处理第6个字符 'P':
密文字符'P'在字母表中的位置是: 18
生成密钥流值:
1. 移动A王牌下移一位
2. 移动B王牌下移两位
3. 以王牌为界进行三切
4. 根据底牌进行计数切牌
生成的密钥流值是: 45
解密计算: (18 - 45) % 26 = 25
解密后的字符是: I

处理第7个字符 'J':
密文字符'J'在字母表中的位置是: 0
生成密钥流值:
1. 移动A王牌下移一位
2. 移动B王牌下移两位
3. 以王牌为界进行三切
4. 根据底牌进行计数切牌
生成的密钥流值是: 45
解密计算: (0 - 45) % 26 = 7
解密后的字符是: C

最终解密结果: PYRITIC


Prompt:
You are an excellent cipher encoder, please refer to the following examples and information to encode the plaintext.
**Encryption Rules:**

- Input:
    - Plaintext: Uppercase letters string without punctuation and spaces.
- Output:
    - Ciphertext: Uppercase letters string.
- Preparation:
    - alphabet = 'ABCDEFGHIKLMNOPQRSTUVWXYZ'(The letter J is excluded)
    - 8 Grids:

        Grid0:

        | P | H | I | L | S |
        | A | B | C | D | E |
        | F | G | K | M | N |
        | O | Q | R | T | U |
        | V | W | X | Y | Z |

        Grid1:

        | V | W | X | Y | Z |
        | P | H | I | L | S |
        | A | B | C | D | E |
        | F | G | K | M | N |
        | O | Q | R | T | U |

        Grid2:

        | O | Q | R | T | U |
        | V | W | X | Y | Z |
        | P | H | I | L | S |
        | A | B | C | D | E |
        | F | G | K | M | N |

        Grid3:

        | F | G | K | M | N |
        | O | Q | R | T | U |
        | V | W | X | Y | Z |
        | P | H | I | L | S |
        | A | B | C | D | E |

        Grid4:

        | A | B | C | D | E |
        | F | G | K | M | N |
        | O | Q | R | T | U |
        | V | W | X | Y | Z |
        | P | H | I | L | S |

        Grid5:

        | P | H | I | L | S |
        | A | B | C | D | E |
        | F | G | K | M | N |
        | O | Q | R | T | U |
        | V | W | X | Y | Z |

        Grid6:

        | V | W | X | Y | Z |
        | P | H | I | L | S |
        | A | B | C | D | E |
        | F | G | K | M | N |
        | O | Q | R | T | U |

        Grid7:

        | O | Q | R | T | U |
        | V | W | X | Y | Z |
        | P | H | I | L | S |
        | A | B | C | D | E |
        | F | G | K | M | N |

- Encryption Steps:
    - The plaintext is grouped into blocks of 5 characters, numbered from 0.
    - For blocks of 5 characters:
        - To use the grid is determined by calculating grid_index = (i // 5) % 8, where i is the number. Integer division operator // divides the number on the left by the number on the right, rounding down the result. That is, it removes the fractional part and keeps only the integer part!This ensures that the grid is recycled.
        - For each character in the current block:
            - If the character is "J", it is not encrypted and is appended directly to the encrypted block.
            - Otherwise, find the position of the character in the current grid. Then move one grid position to the lower right (row+1,col+1) (or continue on the opposite side of the corresponding boundary if it crosses the boundary), and get the letter at the position after the move as the encrypted letter.
            - Add the encrypted letter to the encryption block.
    - After processing all the blocks, connect the encrypted blocks to form the final encrypted message.

plain text: nonculpability encrypt to cipher text: OWOMVEBGKDEDZS
plain text: nonculpability encrypt to cipher text: OWOMVEBGKDEDZS
plain text: nervii encrypt to cipher text: ? Please complete it step by step, ensuring the process is detailed and rigorous, and the result is correct.

Let's think step by step and output the final answer with an example markdown formatting:
Final-answer: ```text
BTWTBIGTKTGBGIKHGTBTBEME
```

Input: NERVII

Output:
处理后的输入文本: NERVII

处理第0个块: NERVI
使用第0号网格
字母N在位置(2,4)，向右下移动一格得到O
字母E在位置(1,4)，向右下移动一格得到F
字母R在位置(3,2)，向右下移动一格得到Y
字母V在位置(4,0)，向右下移动一格得到H
字母I在位置(0,2)，向右下移动一格得到D
加密后的块: OFYHD

处理第1个块: I
使用第1号网格
字母I在位置(1,2)，向右下移动一格得到D
加密后的块: D

最终加密结果: OFYHDD


Prompt:
By your deep knowledge in cryptography, your work is to analyze and apply the encryption techniques in the examples, ensuring the security of information during transmission.
**Encryption Rules:**

- Input:
    - Plaintext: Uppercase letters string without punctuation and spaces.
- Output:
    - Ciphertext: Uppercase letters string.
- Preparation:
    - alphabet = 'ABCDEFGHIKLMNOPQRSTUVWXYZ'(The letter J is excluded)
    - 8 Grids:

        Grid0:

        | P | H | I | L | S |
        | A | B | C | D | E |
        | F | G | K | M | N |
        | O | Q | R | T | U |
        | V | W | X | Y | Z |

        Grid1:

        | V | W | X | Y | Z |
        | P | H | I | L | S |
        | A | B | C | D | E |
        | F | G | K | M | N |
        | O | Q | R | T | U |

        Grid2:

        | O | Q | R | T | U |
        | V | W | X | Y | Z |
        | P | H | I | L | S |
        | A | B | C | D | E |
        | F | G | K | M | N |

        Grid3:

        | F | G | K | M | N |
        | O | Q | R | T | U |
        | V | W | X | Y | Z |
        | P | H | I | L | S |
        | A | B | C | D | E |

        Grid4:

        | A | B | C | D | E |
        | F | G | K | M | N |
        | O | Q | R | T | U |
        | V | W | X | Y | Z |
        | P | H | I | L | S |

        Grid5:

        | P | H | I | L | S |
        | A | B | C | D | E |
        | F | G | K | M | N |
        | O | Q | R | T | U |
        | V | W | X | Y | Z |

        Grid6:

        | V | W | X | Y | Z |
        | P | H | I | L | S |
        | A | B | C | D | E |
        | F | G | K | M | N |
        | O | Q | R | T | U |

        Grid7:

        | O | Q | R | T | U |
        | V | W | X | Y | Z |
        | P | H | I | L | S |
        | A | B | C | D | E |
        | F | G | K | M | N |

- Encryption Steps:
    - The plaintext is grouped into blocks of 5 characters, numbered from 0.
    - For blocks of 5 characters:
        - To use the grid is determined by calculating grid_index = (i // 5) % 8, where i is the number. Integer division operator // divides the number on the left by the number on the right, rounding down the result. That is, it removes the fractional part and keeps only the integer part!This ensures that the grid is recycled.
        - For each character in the current block:
            - If the character is "J", it is not encrypted and is appended directly to the encrypted block.
            - Otherwise, find the position of the character in the current grid. Then move one grid position to the lower right (row+1,col+1) (or continue on the opposite side of the corresponding boundary if it crosses the boundary), and get the letter at the position after the move as the encrypted letter.
            - Add the encrypted letter to the encryption block.
    - After processing all the blocks, connect the encrypted blocks to form the final encrypted message.

clear text: pyritic encrypt to cipher text: BSYDZDM
clear text: pyritic encrypt to cipher text: BSYDZDM
clear text: pyritic encrypt to cipher text: BSYDZDM
clear text: nonculpability encrypt to cipher text is: ? step by step

Let's think step by step and output the final answer with an example markdown formatting:
Final-answer: ```text
BTWTBIGTKTGBGIKHGTBTBEME
```

Input: NONCULPABILITY

Output:
处理后的输入文本: NONCULPABILITY

处理第0个块: NONCU
使用第0号网格
字母N在位置(2,4)，向右下移动一格得到O
字母O在位置(3,0)，向右下移动一格得到W
字母N在位置(2,4)，向右下移动一格得到O
字母C在位置(1,2)，向右下移动一格得到M
字母U在位置(3,4)，向右下移动一格得到V
加密后的块: OWOMV

处理第1个块: LPABI
使用第1号网格
字母L在位置(1,3)，向右下移动一格得到E
字母P在位置(1,0)，向右下移动一格得到B
字母A在位置(2,0)，向右下移动一格得到G
字母B在位置(2,1)，向右下移动一格得到K
字母I在位置(1,2)，向右下移动一格得到D
加密后的块: EBGKD

处理第2个块: LITY
使用第2号网格
字母L在位置(2,3)，向右下移动一格得到E
字母I在位置(2,2)，向右下移动一格得到D
字母T在位置(0,3)，向右下移动一格得到Z
字母Y在位置(1,3)，向右下移动一格得到S
加密后的块: EDZS

最终加密结果: OWOMVEBGKDEDZS


Prompt:
By your deep knowledge in cryptography, your work is to analyze and apply the encryption techniques in the examples, ensuring the security of information during transmission.
**Encryption Rules:**

- Input:
    - Plaintext: Uppercase letters string without punctuation and spaces.
- Output:
    - Ciphertext: Uppercase letters string.
- Preparation:
    - alphabet = 'ABCDEFGHIKLMNOPQRSTUVWXYZ'(The letter J is excluded)
    - 8 Grids:

        Grid0:

        | P | H | I | L | S |
        | A | B | C | D | E |
        | F | G | K | M | N |
        | O | Q | R | T | U |
        | V | W | X | Y | Z |

        Grid1:

        | V | W | X | Y | Z |
        | P | H | I | L | S |
        | A | B | C | D | E |
        | F | G | K | M | N |
        | O | Q | R | T | U |

        Grid2:

        | O | Q | R | T | U |
        | V | W | X | Y | Z |
        | P | H | I | L | S |
        | A | B | C | D | E |
        | F | G | K | M | N |

        Grid3:

        | F | G | K | M | N |
        | O | Q | R | T | U |
        | V | W | X | Y | Z |
        | P | H | I | L | S |
        | A | B | C | D | E |

        Grid4:

        | A | B | C | D | E |
        | F | G | K | M | N |
        | O | Q | R | T | U |
        | V | W | X | Y | Z |
        | P | H | I | L | S |

        Grid5:

        | P | H | I | L | S |
        | A | B | C | D | E |
        | F | G | K | M | N |
        | O | Q | R | T | U |
        | V | W | X | Y | Z |

        Grid6:

        | V | W | X | Y | Z |
        | P | H | I | L | S |
        | A | B | C | D | E |
        | F | G | K | M | N |
        | O | Q | R | T | U |

        Grid7:

        | O | Q | R | T | U |
        | V | W | X | Y | Z |
        | P | H | I | L | S |
        | A | B | C | D | E |
        | F | G | K | M | N |

- Encryption Steps:
    - The plaintext is grouped into blocks of 5 characters, numbered from 0.
    - For blocks of 5 characters:
        - To use the grid is determined by calculating grid_index = (i // 5) % 8, where i is the number. Integer division operator // divides the number on the left by the number on the right, rounding down the result. That is, it removes the fractional part and keeps only the integer part!This ensures that the grid is recycled.
        - For each character in the current block:
            - If the character is "J", it is not encrypted and is appended directly to the encrypted block.
            - Otherwise, find the position of the character in the current grid. Then move one grid position to the lower right (row+1,col+1) (or continue on the opposite side of the corresponding boundary if it crosses the boundary), and get the letter at the position after the move as the encrypted letter.
            - Add the encrypted letter to the encryption block.
    - After processing all the blocks, connect the encrypted blocks to form the final encrypted message.

clear text: nonculpability encrypt into encoded text: OWOMVEBGKDEDZS
clear text: nervii encrypt into encoded text: OFYHDD
clear text: nervii encrypt into encoded text: OFYHDD
clear text: pyritic encrypt into encoded text: ? Please plan the process and implement it step by step, ensuring that each step is carefully checked to ultimately achieve the desired outcome.

Let's think step by step and output the final answer with an example markdown formatting:
Final-answer: ```text
BTWTBIGTKTGBGIKHGTBTBEME
```

Input: PYRITIC

Output:
处理后的输入文本: PYRITIC

处理第0个块: PYRIT
使用第0号网格
字母P在位置(0,0)，向右下移动一格得到B
字母Y在位置(4,3)，向右下移动一格得到S
字母R在位置(3,2)，向右下移动一格得到Y
字母I在位置(0,2)，向右下移动一格得到D
字母T在位置(3,3)，向右下移动一格得到Z
加密后的块: BSYDZ

处理第1个块: IC
使用第1号网格
字母I在位置(1,2)，向右下移动一格得到D
字母C在位置(2,2)，向右下移动一格得到M
加密后的块: DM

最终加密结果: BSYDZDM


Prompt:
As an encoder, your task is to use the encryption algorithm described in the examples to encrypt the plaintext.
clear text: pyritic secret key: {'shift': 1} encrypt into cipher text: qzsjujd
clear text: pyritic secret key: {'shift': 1} encrypt into cipher text: qzsjujd
clear text: nonculpability secret key: {'shift': 15} encrypt into cipher text: cdcrjaepqxaxin
clear text: nonculpability secret key: {'shift': 15} encrypt into cipher text: cdcrjaepqxaxin
clear text: nervii secret key: {'shift': 7} encrypt into cipher text is: ?

Let's think step by step and output the final answer with an example markdown formatting:
Final-answer: ```text
BTWTBIGTKTGBGIKHGTBTBEME
```

Input: nervii

Output:
开始编码过程，文本为: nervii，位移数为: 7
处理字符: n
字符 n 向后移动 7 位，变为 u
处理字符: e
字符 e 向后移动 7 位，变为 l
处理字符: r
字符 r 向后移动 7 位，变为 y
处理字符: v
字符 v 向后移动 7 位，变为 c
处理字符: i
字符 i 向后移动 7 位，变为 p
处理字符: i
字符 i 向后移动 7 位，变为 p
编码完成，结果为: ulycpp


Prompt:
You are an excellent cipher encoder, please refer to the following examples and information to encode the plaintext.
clear text: pyritic secret key: {'shift': 1} encode into encrypted text: qzsjujd
clear text: pyritic secret key: {'shift': 1} encode into encrypted text: qzsjujd
clear text: pyritic secret key: {'shift': 1} encode into encrypted text: qzsjujd
clear text: nervii secret key: {'shift': 7} encode into encrypted text: ulycpp
clear text: pyritic secret key: {'shift': 1} encode into encrypted text: qzsjujd
clear text: nonculpability secret key: {'shift': 15} encode into encrypted text: ? Please proceed carefully according to the steps, ensuring that every detail in the process is not overlooked, to achieve an accurate and error-free goal.

Let's think step by step and output the final answer with an example markdown formatting:
Final-answer: ```text
BTWTBIGTKTGBGIKHGTBTBEME
```

Input: nonculpability

Output:
开始编码过程，文本为: nonculpability，位移数为: 15
处理字符: n
字符 n 向后移动 15 位，变为 c
处理字符: o
字符 o 向后移动 15 位，变为 d
处理字符: n
字符 n 向后移动 15 位，变为 c
处理字符: c
字符 c 向后移动 15 位，变为 r
处理字符: u
字符 u 向后移动 15 位，变为 j
处理字符: l
字符 l 向后移动 15 位，变为 a
处理字符: p
字符 p 向后移动 15 位，变为 e
处理字符: a
字符 a 向后移动 15 位，变为 p
处理字符: b
字符 b 向后移动 15 位，变为 q
处理字符: i
字符 i 向后移动 15 位，变为 x
处理字符: l
字符 l 向后移动 15 位，变为 a
处理字符: i
字符 i 向后移动 15 位，变为 x
处理字符: t
字符 t 向后移动 15 位，变为 i
处理字符: y
字符 y 向后移动 15 位，变为 n
编码完成，结果为: cdcrjaepqxaxin


Prompt:
As a specialist in cryptography, your job is to analyze the encryption algorithm in the examples and implement the encryption process on the data.
plain text: nonculpability secret key: {'shift': 15} encrypt to cipher text: cdcrjaepqxaxin
plain text: nonculpability secret key: {'shift': 15} encrypt to cipher text: cdcrjaepqxaxin
plain text: nervii secret key: {'shift': 7} encrypt to cipher text: ulycpp
plain text: nonculpability secret key: {'shift': 15} encrypt to cipher text: cdcrjaepqxaxin
plain text: nervii secret key: {'shift': 7} encrypt to cipher text: ulycpp
plain text: pyritic secret key: {'shift': 1} encrypt to cipher text: ? Please complete it step by step, ensuring the process is detailed and rigorous, and the result is correct.

Let's think step by step and output the final answer with an example markdown formatting:
Final-answer: ```text
BTWTBIGTKTGBGIKHGTBTBEME
```

Input: pyritic

Output:
开始编码过程，文本为: pyritic，位移数为: 1
处理字符: p
字符 p 向后移动 1 位，变为 q
处理字符: y
字符 y 向后移动 1 位，变为 z
处理字符: r
字符 r 向后移动 1 位，变为 s
处理字符: i
字符 i 向后移动 1 位，变为 j
处理字符: t
字符 t 向后移动 1 位，变为 u
处理字符: i
字符 i 向后移动 1 位，变为 j
处理字符: c
字符 c 向后移动 1 位，变为 d
编码完成，结果为: qzsjujd


Prompt:
作为密码学领域的专家，您需要运用您的专业知识，分析案例中的加密算法，并对数据实施解密处理。
请根据加密算法和解密算法对密码进行解密
加密算法:
Tap Code 是一种简单的密码技术，每个字母由一个点模式表示。模式通过计算行和列中的点数来编码。例如，字母 'A' 由第一行和第一列的一个点表示。
解密算法:
要解码 Tap Code，需要计算每一行和每一列中的点数。通过查找表格中的位置，可以确定对应的字母。
暗码: ... .....  ..... ....  .... ..  .. ....  .... ....  .. ....  . ... 解码为明文: pyritic
暗码: ... ...  . .....  .... ..  ..... .  .. ....  .. .... 解码为明文: ?

Let's think step by step and output the final answer with an example markdown formatting:
Final-answer: ```text
BTWTBIGTKTGBGIKHGTBTBEME
```

Output:
步骤 1：正在解码 '... ...'，它表示第 3 行第 3 列，对应的字母是 'N'
步骤 2：正在解码 '. .....'，它表示第 1 行第 5 列，对应的字母是 'E'
步骤 3：正在解码 '.... ..'，它表示第 4 行第 2 列，对应的字母是 'R'
步骤 4：正在解码 '..... .'，它表示第 5 行第 1 列，对应的字母是 'V'
步骤 5：正在解码 '.. ....'，它表示第 2 行第 4 列，对应的字母是 'I'
步骤 6：正在解码 '.. ....'，它表示第 2 行第 4 列，对应的字母是 'I'
最终步骤：解码完成，解码后的消息是：NERVII


Prompt:
您的任务是使用正确的算法将看似无意义的密文转换回可读的原始信息，确保信息的准确性和完整性。
请根据加密算法和解密算法对密码进行解密
加密算法:
Tap Code 是一种简单的密码技术，每个字母由一个点模式表示。模式通过计算行和列中的点数来编码。例如，字母 'A' 由第一行和第一列的一个点表示。
解密算法:
要解码 Tap Code，需要计算每一行和每一列中的点数。通过查找表格中的位置，可以确定对应的字母。
加密信息: ... ...  ... ....  ... ...  . ...  .... .....  ... .  ... .....  . .  . ..  .. ....  ... .  .. ....  .... ....  ..... .... 解密为原文是什么？ 一步一步完成

Let's think step by step and output the final answer with an example markdown formatting:
Final-answer: ```text
BTWTBIGTKTGBGIKHGTBTBEME
```

Output:
步骤 1：正在解码 '... ...'，它表示第 3 行第 3 列，对应的字母是 'N'
步骤 2：正在解码 '... ....'，它表示第 3 行第 4 列，对应的字母是 'O'
步骤 3：正在解码 '... ...'，它表示第 3 行第 3 列，对应的字母是 'N'
步骤 4：正在解码 '. ...'，它表示第 1 行第 3 列，对应的字母是 'C'
步骤 5：正在解码 '.... .....'，它表示第 4 行第 5 列，对应的字母是 'U'
步骤 6：正在解码 '... .'，它表示第 3 行第 1 列，对应的字母是 'L'
步骤 7：正在解码 '... .....'，它表示第 3 行第 5 列，对应的字母是 'P'
步骤 8：正在解码 '. .'，它表示第 1 行第 1 列，对应的字母是 'A'
步骤 9：正在解码 '. ..'，它表示第 1 行第 2 列，对应的字母是 'B'
步骤 10：正在解码 '.. ....'，它表示第 2 行第 4 列，对应的字母是 'I'
步骤 11：正在解码 '... .'，它表示第 3 行第 1 列，对应的字母是 'L'
步骤 12：正在解码 '.. ....'，它表示第 2 行第 4 列，对应的字母是 'I'
步骤 13：正在解码 '.... ....'，它表示第 4 行第 4 列，对应的字母是 'T'
步骤 14：正在解码 '..... ....'，它表示第 5 行第 4 列，对应的字母是 'Y'
最终步骤：解码完成，解码后的消息是：NONCULPABILITY


Prompt:
请根据加密算法和解密算法对密码进行解密
加密算法:
Tap Code 是一种简单的密码技术，每个字母由一个点模式表示。模式通过计算行和列中的点数来编码。例如，字母 'A' 由第一行和第一列的一个点表示。
解密算法:
要解码 Tap Code，需要计算每一行和每一列中的点数。通过查找表格中的位置，可以确定对应的字母。
暗码: ... .....  ..... ....  .... ..  .. ....  .... ....  .. ....  . ... 解密成初始文本是什么？ 请细心地依照步骤行动，确保过程中的每个细节都不被忽视，以达成准确无误的目标。

Let's think step by step and output the final answer with an example markdown formatting:
Final-answer: ```text
BTWTBIGTKTGBGIKHGTBTBEME
```

Output:
步骤 1：正在解码 '... .....'，它表示第 3 行第 5 列，对应的字母是 'P'
步骤 2：正在解码 '..... ....'，它表示第 5 行第 4 列，对应的字母是 'Y'
步骤 3：正在解码 '.... ..'，它表示第 4 行第 2 列，对应的字母是 'R'
步骤 4：正在解码 '.. ....'，它表示第 2 行第 4 列，对应的字母是 'I'
步骤 5：正在解码 '.... ....'，它表示第 4 行第 4 列，对应的字母是 'T'
步骤 6：正在解码 '.. ....'，它表示第 2 行第 4 列，对应的字母是 'I'
步骤 7：正在解码 '. ...'，它表示第 1 行第 3 列，对应的字母是 'C'
最终步骤：解码完成，解码后的消息是：PYRITIC


Prompt:
You are an excellent cipher decoder, please refer to the following examples and information to decode the ciphertext.
encrypted text: cdcrjaepqxaxin secret key: {'shift': 15} decrypt into clear text: nonculpability
encrypted text: cdcrjaepqxaxin secret key: {'shift': 15} decrypt into clear text: nonculpability
encrypted text: cdcrjaepqxaxin secret key: {'shift': 15} decrypt into clear text: nonculpability
encrypted text: qzsjujd secret key: {'shift': 1} decrypt into clear text: pyritic
encrypted text: ulycpp secret key: {'shift': 7} decrypt into clear text: ?  Carefully complete each step.

Let's think step by step and output the final answer with an example markdown formatting:
Final-answer: ```text
BTWTBIGTKTGBGIKHGTBTBEME
```

Input: ulycpp

Output:
开始解码过程，文本为: ulycpp，位移数为: 7
处理字符: u
字符 u 向前移动 7 位，变为 n
处理字符: l
字符 l 向前移动 7 位，变为 e
处理字符: y
字符 y 向前移动 7 位，变为 r
处理字符: c
字符 c 向前移动 7 位，变为 v
处理字符: p
字符 p 向前移动 7 位，变为 i
处理字符: p
字符 p 向前移动 7 位，变为 i
解码完成，结果为: nervii


Prompt:
encoded text: qzsjujd extra parameter: {'shift': 1} decode into clear text: pyritic
encoded text: qzsjujd extra parameter: {'shift': 1} decode into clear text: pyritic
encoded text: cdcrjaepqxaxin extra parameter: {'shift': 15} decode into clear text is: ? Please complete it step by step, ensuring the process is detailed and rigorous, and the result is correct.

Let's think step by step and output the final answer with an example markdown formatting:
Final-answer: ```text
BTWTBIGTKTGBGIKHGTBTBEME
```

Input: cdcrjaepqxaxin

Output:
开始解码过程，文本为: cdcrjaepqxaxin，位移数为: 15
处理字符: c
字符 c 向前移动 15 位，变为 n
处理字符: d
字符 d 向前移动 15 位，变为 o
处理字符: c
字符 c 向前移动 15 位，变为 n
处理字符: r
字符 r 向前移动 15 位，变为 c
处理字符: j
字符 j 向前移动 15 位，变为 u
处理字符: a
字符 a 向前移动 15 位，变为 l
处理字符: e
字符 e 向前移动 15 位，变为 p
处理字符: p
字符 p 向前移动 15 位，变为 a
处理字符: q
字符 q 向前移动 15 位，变为 b
处理字符: x
字符 x 向前移动 15 位，变为 i
处理字符: a
字符 a 向前移动 15 位，变为 l
处理字符: x
字符 x 向前移动 15 位，变为 i
处理字符: i
字符 i 向前移动 15 位，变为 t
处理字符: n
字符 n 向前移动 15 位，变为 y
解码完成，结果为: nonculpability


Prompt:
By your deep knowledge in cryptography, your work is to analyze and apply the encryption techniques in the examples, ensuring the security of information during transmission.
cipher text: cdcrjaepqxaxin secret key: {'shift': 15} decrypt into plain text: nonculpability
cipher text: cdcrjaepqxaxin secret key: {'shift': 15} decrypt into plain text: nonculpability
cipher text: cdcrjaepqxaxin secret key: {'shift': 15} decrypt into plain text: nonculpability
cipher text: cdcrjaepqxaxin secret key: {'shift': 15} decrypt into plain text: nonculpability
cipher text: qzsjujd secret key: {'shift': 1} decrypt into plain text is: ? Please complete it step by step, formulate a reasonable problem-solving plan, and strictly adhere to it.

Let's think step by step and output the final answer with an example markdown formatting:
Final-answer: ```text
BTWTBIGTKTGBGIKHGTBTBEME
```

Input: qzsjujd

Output:
开始解码过程，文本为: qzsjujd，位移数为: 1
处理字符: q
字符 q 向前移动 1 位，变为 p
处理字符: z
字符 z 向前移动 1 位，变为 y
处理字符: s
字符 s 向前移动 1 位，变为 r
处理字符: j
字符 j 向前移动 1 位，变为 i
处理字符: u
字符 u 向前移动 1 位，变为 t
处理字符: j
字符 j 向前移动 1 位，变为 i
处理字符: d
字符 d 向前移动 1 位，变为 c
解码完成，结果为: pyritic


Prompt:
凭借您在密码学方面的深厚造诣，您的工作是分析并应用案例中的加密技术，确保信息在传输过程中不会被非法截获。
请根据加密算法和解密算法对密码进行解密
加密算法:
加密规则：
1. 将输入文本中的每个字母向后移动指定的位数（shift）。
2. 输出加密后的文本。
解密算法:
解码规则：
1. 将输入文本中的每个字母向前移动指定的位数（shift）。
2. 输出解密后的文本。
暗文: qzsjujd 密钥: {'shift': 1} 解密成初始文本: pyritic
暗文: qzsjujd 密钥: {'shift': 1} 解密成初始文本: pyritic
暗文: qzsjujd 密钥: {'shift': 1} 解密成初始文本: pyritic
暗文: ulycpp 密钥: {'shift': 7} 解密成初始文本: ? 请细心地依照步骤行动，确保过程中的每个细节都不被忽视，以达成准确无误的目标。

Let's think step by step and output the final answer with an example markdown formatting:
Final-answer: ```text
BTWTBIGTKTGBGIKHGTBTBEME
```

Input: ulycpp

Output:
开始解码过程，文本为: ulycpp，位移数为: 7
处理字符: u
字符 u 向前移动 7 位，变为 n
处理字符: l
字符 l 向前移动 7 位，变为 e
处理字符: y
字符 y 向前移动 7 位，变为 r
处理字符: c
字符 c 向前移动 7 位，变为 v
处理字符: p
字符 p 向前移动 7 位，变为 i
处理字符: p
字符 p 向前移动 7 位，变为 i
解码完成，结果为: nervii


Prompt:
凭借您在密码学方面的深厚造诣，您的工作是分析并应用案例中的加密技术，确保信息在传输过程中不会被非法截获。
请根据加密算法和解密算法对密码进行解密
加密算法:
加密规则：
1. 将输入文本中的每个字母向后移动指定的位数（shift）。
2. 输出加密后的文本。
解密算法:
解码规则：
1. 将输入文本中的每个字母向前移动指定的位数（shift）。
2. 输出解密后的文本。
暗码: ulycpp 额外参数: {'shift': 7} 解密成初始文本: nervii
暗码: qzsjujd 额外参数: {'shift': 1} 解密成初始文本: pyritic
暗码: ulycpp 额外参数: {'shift': 7} 解密成初始文本: nervii
暗码: cdcrjaepqxaxin 额外参数: {'shift': 15} 解密成初始文本是什么？ 请规划流程一步步实施，确保每一步都经过仔细检查，最终达到预期效果。

Let's think step by step and output the final answer with an example markdown formatting:
Final-answer: ```text
BTWTBIGTKTGBGIKHGTBTBEME
```

Input: cdcrjaepqxaxin

Output:
开始解码过程，文本为: cdcrjaepqxaxin，位移数为: 15
处理字符: c
字符 c 向前移动 15 位，变为 n
处理字符: d
字符 d 向前移动 15 位，变为 o
处理字符: c
字符 c 向前移动 15 位，变为 n
处理字符: r
字符 r 向前移动 15 位，变为 c
处理字符: j
字符 j 向前移动 15 位，变为 u
处理字符: a
字符 a 向前移动 15 位，变为 l
处理字符: e
字符 e 向前移动 15 位，变为 p
处理字符: p
字符 p 向前移动 15 位，变为 a
处理字符: q
字符 q 向前移动 15 位，变为 b
处理字符: x
字符 x 向前移动 15 位，变为 i
处理字符: a
字符 a 向前移动 15 位，变为 l
处理字符: x
字符 x 向前移动 15 位，变为 i
处理字符: i
字符 i 向前移动 15 位，变为 t
处理字符: n
字符 n 向前移动 15 位，变为 y
解码完成，结果为: nonculpability


Prompt:
您的任务是使用正确的算法将看似无意义的密文转换回可读的原始信息，确保信息的准确性和完整性。
请根据加密算法和解密算法对密码进行解密
加密算法:
加密规则：
1. 将输入文本中的每个字母向后移动指定的位数（shift）。
2. 输出加密后的文本。
解密算法:
解码规则：
1. 将输入文本中的每个字母向前移动指定的位数（shift）。
2. 输出解密后的文本。
加密文本: qzsjujd 额外参数: {'shift': 1} 解码为非加密信息是什么？ 请细心地依照步骤行动，确保过程中的每个细节都不被忽视，以达成准确无误的目标。

Let's think step by step and output the final answer with an example markdown formatting:
Final-answer: ```text
BTWTBIGTKTGBGIKHGTBTBEME
```

Input: qzsjujd

Output:
开始解码过程，文本为: qzsjujd，位移数为: 1
处理字符: q
字符 q 向前移动 1 位，变为 p
处理字符: z
字符 z 向前移动 1 位，变为 y
处理字符: s
字符 s 向前移动 1 位，变为 r
处理字符: j
字符 j 向前移动 1 位，变为 i
处理字符: u
字符 u 向前移动 1 位，变为 t
处理字符: j
字符 j 向前移动 1 位，变为 i
处理字符: d
字符 d 向前移动 1 位，变为 c
解码完成，结果为: pyritic


Prompt:
Your primary responsibility is to use your professional skills to encode the provided plaintext using the correct algorithm and ensure the security of information during transmission.
**Encryption Rules:**

- Input:
    - Plaintext: Uppercase letters string without punctuation and spaces.
- Output:
    - Ciphertext: Uppercase letters string.
- Preparation:
    - 5x5 grid (There is no letter J.Its row and column coordinates are between 0 and 4.):
        - U  B  I  L  A
        N  T  C  D  E
        F  G  H  K  M
        O  P  Q  R  S
        V  W  X  Y  Z
- Encryption Steps:
    - Remove punctuation, spaces, and the letter J from plaintext.
    - For each letter p in the plaintext, find its corresponding row and column coordinates (between 0 and 4) in the grid. For example, A is (0,4).
    - Arrange all row and column coordinates, read all row coordinates first, then read all column coordinates to form a new coordinate sequence. For example, the original has (0,4)(1,2), now read as 0142.
    - Each time, two numbers are taken out from the new coordinate sequence as new row and column coordinates, and the corresponding letters are found in the grid to form the ciphertext. For example, for 0142, the ciphertext corresponding to (0,1) is B, and the ciphertext corresponding to (4,2) is X. The final ciphertext is BX.

clear text: nervii encrypt into encrypted text is: ?

Let's think step by step and output the final answer with an example markdown formatting:
Final-answer: ```text
BTWTBIGTKTGBGIKHGTBTBEME
```

Input: NERVII

Output:
加密步骤:
1. 去除标点和空格,将J替换为I后的文本: NERVII
2. 生成的5x5矩阵:
['U', 'B', 'I', 'L', 'A']
['N', 'T', 'C', 'D', 'E']
['F', 'G', 'H', 'K', 'M']
['O', 'P', 'Q', 'R', 'S']
['V', 'W', 'X', 'Y', 'Z']
3. 获取每个字母的行列坐标: [1, 1, 3, 4, 0, 0, 0, 4, 3, 0, 2, 2]
4. 最终加密结果: TSUAOH


Prompt:
As an encoder, your task is to use the encryption algorithm described in the examples to encrypt the plaintext.
**Encryption Rules:**

- Input:
    - Plaintext: Uppercase letters string without punctuation and spaces.
- Output:
    - Ciphertext: Uppercase letters string.
- Preparation:
    - 5x5 grid (There is no letter J.Its row and column coordinates are between 0 and 4.):
        - U  B  I  L  A
        N  T  C  D  E
        F  G  H  K  M
        O  P  Q  R  S
        V  W  X  Y  Z
- Encryption Steps:
    - Remove punctuation, spaces, and the letter J from plaintext.
    - For each letter p in the plaintext, find its corresponding row and column coordinates (between 0 and 4) in the grid. For example, A is (0,4).
    - Arrange all row and column coordinates, read all row coordinates first, then read all column coordinates to form a new coordinate sequence. For example, the original has (0,4)(1,2), now read as 0142.
    - Each time, two numbers are taken out from the new coordinate sequence as new row and column coordinates, and the corresponding letters are found in the grid to form the ciphertext. For example, for 0142, the ciphertext corresponding to (0,1) is B, and the ciphertext corresponding to (4,2) is X. The final ciphertext is BX.

plain text: nervii encode to encrypted text: TSUAOH
plain text: pyritic encode to encrypted text: SONTRGH
plain text: pyritic encode to encrypted text: SONTRGH
plain text: nonculpability encode to encrypted text: ? Please proceed carefully according to the steps, ensuring that every detail in the process is not overlooked, to achieve an accurate and error-free goal.

Let's think step by step and output the final answer with an example markdown formatting:
Final-answer: ```text
BTWTBIGTKTGBGIKHGTBTBEME
```

Input: NONCULPABILITY

Output:
加密步骤:
1. 去除标点和空格,将J替换为I后的文本: NONCULPABILITY
2. 生成的5x5矩阵:
['U', 'B', 'I', 'L', 'A']
['N', 'T', 'C', 'D', 'E']
['F', 'G', 'H', 'K', 'M']
['O', 'P', 'Q', 'R', 'S']
['V', 'W', 'X', 'Y', 'Z']
3. 获取每个字母的行列坐标: [1, 3, 1, 1, 0, 0, 3, 0, 0, 0, 0, 0, 1, 4, 0, 0, 0, 2, 0, 3, 1, 4, 1, 2, 3, 2, 1, 3]
4. 最终加密结果: DTUOUUEUILECQD


Prompt:
Your primary responsibility is to use your professional skills to encode the provided plaintext using the correct algorithm and ensure the security of information during transmission.
**Encryption Rules:**

- Input:
    - Plaintext: Uppercase letters string without punctuation and spaces.
- Output:
    - Ciphertext: Uppercase letters string.
- Preparation:
    - 5x5 grid (There is no letter J.Its row and column coordinates are between 0 and 4.):
        - U  B  I  L  A
        N  T  C  D  E
        F  G  H  K  M
        O  P  Q  R  S
        V  W  X  Y  Z
- Encryption Steps:
    - Remove punctuation, spaces, and the letter J from plaintext.
    - For each letter p in the plaintext, find its corresponding row and column coordinates (between 0 and 4) in the grid. For example, A is (0,4).
    - Arrange all row and column coordinates, read all row coordinates first, then read all column coordinates to form a new coordinate sequence. For example, the original has (0,4)(1,2), now read as 0142.
    - Each time, two numbers are taken out from the new coordinate sequence as new row and column coordinates, and the corresponding letters are found in the grid to form the ciphertext. For example, for 0142, the ciphertext corresponding to (0,1) is B, and the ciphertext corresponding to (4,2) is X. The final ciphertext is BX.

clear text: nervii encrypt into cipher text: TSUAOH
clear text: nervii encrypt into cipher text: TSUAOH
clear text: pyritic encrypt into cipher text is: ? Please complete it step by step, formulate a reasonable problem-solving plan, and strictly adhere to it.

Let's think step by step and output the final answer with an example markdown formatting:
Final-answer: ```text
BTWTBIGTKTGBGIKHGTBTBEME
```

Input: PYRITIC

Output:
加密步骤:
1. 去除标点和空格,将J替换为I后的文本: PYRITIC
2. 生成的5x5矩阵:
['U', 'B', 'I', 'L', 'A']
['N', 'T', 'C', 'D', 'E']
['F', 'G', 'H', 'K', 'M']
['O', 'P', 'Q', 'R', 'S']
['V', 'W', 'X', 'Y', 'Z']
3. 获取每个字母的行列坐标: [3, 4, 3, 0, 1, 0, 1, 1, 3, 3, 2, 1, 2, 2]
4. 最终加密结果: SONTRGH


Prompt:
As a specialist in cryptography, your job is to analyze the encryption algorithm in the examples and implement the encryption process on the data.
**Encryption Rules:**

- Input:
    - Plaintext: Uppercase letters string without punctuation and spaces.
- Output:
    - Ciphertext:  A sequence of numbers separated by ,. For example, "y1,y2,…".
- Preparation:
    - k_q_x : 12
- Encryption Steps:
    - For each letter p in the plaintext:
        - Get the decimal number x corresponding to the ascii code of p.
        - Calculate x * k_q_x as the number y of ciphertexts for that letter p.
    - Finally, concatenate all y's and separate them with commas to get the final ciphertext.

clear text: pyritic encrypt to encoded text: 960,1068,984,876,1008,876,804
clear text: pyritic encrypt to encoded text: 960,1068,984,876,1008,876,804
clear text: nonculpability encrypt to encoded text: 936,948,936,804,1020,912,960,780,792,876,912,876,1008,1068
clear text: nervii encrypt to encoded text is: ?

Let's think step by step and output the final answer with an example markdown formatting:
Final-answer: ```text
BTWTBIGTKTGBGIKHGTBTBEME
```

Input: NERVII

Output:
开始加密过程...
处理后的输入文本: NERVII
使用预设的k_q_x值: 12

处理字符: N
对应的ASCII值: 78
加密后的值: 936

处理字符: E
对应的ASCII值: 69
加密后的值: 828

处理字符: R
对应的ASCII值: 82
加密后的值: 984

处理字符: V
对应的ASCII值: 86
加密后的值: 1032

处理字符: I
对应的ASCII值: 73
加密后的值: 876

处理字符: I
对应的ASCII值: 73
加密后的值: 876

最终加密结果: 936,828,984,1032,876,876


Prompt:
You are an excellent cipher encoder, please refer to the following examples and information to encode the plaintext.
**Encryption Rules:**

- Input:
    - Plaintext: Uppercase letters string without punctuation and spaces.
- Output:
    - Ciphertext:  A sequence of numbers separated by ,. For example, "y1,y2,…".
- Preparation:
    - k_q_x : 12
- Encryption Steps:
    - For each letter p in the plaintext:
        - Get the decimal number x corresponding to the ascii code of p.
        - Calculate x * k_q_x as the number y of ciphertexts for that letter p.
    - Finally, concatenate all y's and separate them with commas to get the final ciphertext.

original information: pyritic encrypt to encrypted text: 960,1068,984,876,1008,876,804
original information: pyritic encrypt to encrypted text: 960,1068,984,876,1008,876,804
original information: pyritic encrypt to encrypted text: 960,1068,984,876,1008,876,804
original information: nonculpability encrypt to encrypted text: ? Please complete it step by step, formulate a reasonable problem-solving plan, and strictly adhere to it.

Let's think step by step and output the final answer with an example markdown formatting:
Final-answer: ```text
BTWTBIGTKTGBGIKHGTBTBEME
```

Input: NONCULPABILITY

Output:
开始加密过程...
处理后的输入文本: NONCULPABILITY
使用预设的k_q_x值: 12

处理字符: N
对应的ASCII值: 78
加密后的值: 936

处理字符: O
对应的ASCII值: 79
加密后的值: 948

处理字符: N
对应的ASCII值: 78
加密后的值: 936

处理字符: C
对应的ASCII值: 67
加密后的值: 804

处理字符: U
对应的ASCII值: 85
加密后的值: 1020

处理字符: L
对应的ASCII值: 76
加密后的值: 912

处理字符: P
对应的ASCII值: 80
加密后的值: 960

处理字符: A
对应的ASCII值: 65
加密后的值: 780

处理字符: B
对应的ASCII值: 66
加密后的值: 792

处理字符: I
对应的ASCII值: 73
加密后的值: 876

处理字符: L
对应的ASCII值: 76
加密后的值: 912

处理字符: I
对应的ASCII值: 73
加密后的值: 876

处理字符: T
对应的ASCII值: 84
加密后的值: 1008

处理字符: Y
对应的ASCII值: 89
加密后的值: 1068

最终加密结果: 936,948,936,804,1020,912,960,780,792,876,912,876,1008,1068


Prompt:
**Encryption Rules:**

- Input:
    - Plaintext: Uppercase letters string without punctuation and spaces.
- Output:
    - Ciphertext:  A sequence of numbers separated by ,. For example, "y1,y2,…".
- Preparation:
    - k_q_x : 12
- Encryption Steps:
    - For each letter p in the plaintext:
        - Get the decimal number x corresponding to the ascii code of p.
        - Calculate x * k_q_x as the number y of ciphertexts for that letter p.
    - Finally, concatenate all y's and separate them with commas to get the final ciphertext.

original information: nervii encrypt into cipher text: 936,828,984,1032,876,876
original information: pyritic encrypt into cipher text: ? step by step

Let's think step by step and output the final answer with an example markdown formatting:
Final-answer: ```text
BTWTBIGTKTGBGIKHGTBTBEME
```

Input: PYRITIC

Output:
开始加密过程...
处理后的输入文本: PYRITIC
使用预设的k_q_x值: 12

处理字符: P
对应的ASCII值: 80
加密后的值: 960

处理字符: Y
对应的ASCII值: 89
加密后的值: 1068

处理字符: R
对应的ASCII值: 82
加密后的值: 984

处理字符: I
对应的ASCII值: 73
加密后的值: 876

处理字符: T
对应的ASCII值: 84
加密后的值: 1008

处理字符: I
对应的ASCII值: 73
加密后的值: 876

处理字符: C
对应的ASCII值: 67
加密后的值: 804

最终加密结果: 960,1068,984,876,1008,876,804


Prompt:
作为密码学领域的专家，您需要运用您的专业知识，分析案例中的加密算法，并对数据实施解密处理。
请根据加密算法和解密算法对密码进行解密
加密算法:

加密规则:
- 输入:
    - 非加密信息: 不含标点和空格的大写字母字符串
- 输出:
    - 加密文本: 不含标点和空格的字符串
- 准备:
    - 行数: 5
- 加密步骤:
    - 在第一行第一个位置填入第一个非加密信息字母
    - 两种填充方式:
        - 向下填充: 在第一行填入"#"(除去第一列，因为第一个位置已经填入非加密信息字母)，然后从第二行到最后一行(第五行)向下填充非加密信息
        - 向上填充: 从最后一行(第五行)到第二行向上填充非加密信息，然后在第一行填入"#"
    - 对于非加密信息中的每个字母(除了已经填在第一个位置的第一个字母)，先进行向下填充，填满一列，然后转向上填充，再转向下填充，如此交替进行，直到所有字母都被填入
    - 填写完成后，逐行读取，读取每行内容后都添加一个*号，标记行的结束；然后读取第二行内容，依此类推，读取所有行，形成最终加密文本

解密算法:

解密规则:
- 输入:
    - 加密文本: 不含标点和空格的字符串
- 输出:
    - 非加密信息: 不含标点和空格的大写字母字符串
- 准备:
    - 行数: 5
- 解密步骤:
    - 根据加密文本中的*(不包括*号)，可以分成五组，依次填入五行。得到恢复的五行数据
    - 然后按照先向下读再向上读交替的方式读取所有列。得到未清理的消息
    - 从未清理的消息中删除#，得到最终非加密信息

加密文本: N*E*R*VI*I#* 解密成为非加密信息是什么？ 请规划流程一步步实施，确保每一步都经过仔细检查，最终达到预期效果。

Let's think step by step and output the final answer with an example markdown formatting:
Final-answer: ```text
BTWTBIGTKTGBGIKHGTBTBEME
```

Input: N*E*R*VI*I#*

Output:
解密步骤开始:
1. 接收到的密文: N*E*R*VI*I#*
2. 根据*号分割密文
  分割得到6个部分
3. 将分割后的内容填入矩阵:
  第1行: N
  第2行: E
  第3行: R
  第4行: VI
  第5行: I#
  第6行:

4. 交替读取矩阵内容:
  向下读取第1列
  向上读取第2列
  向下读取第3列
  向上读取第4列
  向下读取第5列
  向上读取第6列
  向下读取第7列
  向上读取第8列
  向下读取第9列
  向上读取第10列
  向下读取第11列
  向上读取第12列
5. 移除#号获得最终明文
解密结果: NERVII


Prompt:
您的任务是使用正确的算法将看似无意义的密文转换回可读的原始信息，确保信息的准确性和完整性。
请根据加密算法和解密算法对密码进行解密
加密算法:

加密规则:
- 输入:
    - 非加密信息: 不含标点和空格的大写字母字符串
- 输出:
    - 密文: 不含标点和空格的字符串
- 准备:
    - 行数: 5
- 加密步骤:
    - 在第一行第一个位置填入第一个非加密信息字母
    - 两种填充方式:
        - 向下填充: 在第一行填入"#"(除去第一列，因为第一个位置已经填入非加密信息字母)，然后从第二行到最后一行(第五行)向下填充非加密信息
        - 向上填充: 从最后一行(第五行)到第二行向上填充非加密信息，然后在第一行填入"#"
    - 对于非加密信息中的每个字母(除了已经填在第一个位置的第一个字母)，先进行向下填充，填满一列，然后转向上填充，再转向下填充，如此交替进行，直到所有字母都被填入
    - 填写完成后，逐行读取，读取每行内容后都添加一个*号，标记行的结束；然后读取第二行内容，依此类推，读取所有行，形成最终密文

解密算法:

解密规则:
- 输入:
    - 密文: 不含标点和空格的字符串
- 输出:
    - 非加密信息: 不含标点和空格的大写字母字符串
- 准备:
    - 行数: 5
- 解密步骤:
    - 根据密文中的*(不包括*号)，可以分成五组，依次填入五行。得到恢复的五行数据
    - 然后按照先向下读再向上读交替的方式读取所有列。得到未清理的消息
    - 从未清理的消息中删除#，得到最终非加密信息

密文: N*E*R*VI*I#* 解密成非加密信息: nervii
密文: N*E*R*VI*I#* 解密成非加密信息: nervii
密文: N*E*R*VI*I#* 解密成非加密信息: nervii
密文: NB#*OAI*NPL*CLIY*U#T#* 解密成非加密信息是什么？

Let's think step by step and output the final answer with an example markdown formatting:
Final-answer: ```text
BTWTBIGTKTGBGIKHGTBTBEME
```

Input: NB#*OAI*NPL*CLIY*U#T#*

Output:
解密步骤开始:
1. 接收到的密文: NB#*OAI*NPL*CLIY*U#T#*
2. 根据*号分割密文
  分割得到6个部分
3. 将分割后的内容填入矩阵:
  第1行: NB#
  第2行: OAI
  第3行: NPL
  第4行: CLIY
  第5行: U#T#
  第6行:

4. 交替读取矩阵内容:
  向下读取第1列
  向上读取第2列
  向下读取第3列
  向上读取第4列
  向下读取第5列
  向上读取第6列
  向下读取第7列
  向上读取第8列
  向下读取第9列
  向上读取第10列
  向下读取第11列
  向上读取第12列
  向下读取第13列
  向上读取第14列
  向下读取第15列
  向上读取第16列
  向下读取第17列
  向上读取第18列
  向下读取第19列
  向上读取第20列
  向下读取第21列
  向上读取第22列
5. 移除#号获得最终明文
解密结果: NONCULPABILITY


Prompt:
您的主要职责是运用您的专业技能，将提供的密文通过恰当的解密算法恢复成最初的数据形式。
请根据加密算法和解密算法对密码进行解密
加密算法:

加密规则:
- 输入:
    - 原始信息: 不含标点和空格的大写字母字符串
- 输出:
    - 加密文本: 不含标点和空格的字符串
- 准备:
    - 行数: 5
- 加密步骤:
    - 在第一行第一个位置填入第一个原始信息字母
    - 两种填充方式:
        - 向下填充: 在第一行填入"#"(除去第一列，因为第一个位置已经填入原始信息字母)，然后从第二行到最后一行(第五行)向下填充原始信息
        - 向上填充: 从最后一行(第五行)到第二行向上填充原始信息，然后在第一行填入"#"
    - 对于原始信息中的每个字母(除了已经填在第一个位置的第一个字母)，先进行向下填充，填满一列，然后转向上填充，再转向下填充，如此交替进行，直到所有字母都被填入
    - 填写完成后，逐行读取，读取每行内容后都添加一个*号，标记行的结束；然后读取第二行内容，依此类推，读取所有行，形成最终加密文本

解密算法:

解密规则:
- 输入:
    - 加密文本: 不含标点和空格的字符串
- 输出:
    - 原始信息: 不含标点和空格的大写字母字符串
- 准备:
    - 行数: 5
- 解密步骤:
    - 根据加密文本中的*(不包括*号)，可以分成五组，依次填入五行。得到恢复的五行数据
    - 然后按照先向下读再向上读交替的方式读取所有列。得到未清理的消息
    - 从未清理的消息中删除#，得到最终原始信息

加密文本: NB#*OAI*NPL*CLIY*U#T#* 解密为原始信息: nonculpability
加密文本: NB#*OAI*NPL*CLIY*U#T#* 解密为原始信息: nonculpability
加密文本: N*E*R*VI*I#* 解密为原始信息: nervii
加密文本: P*Y*RC*II*T#* 解密为原始信息是什么？ 一步一步完成

Let's think step by step and output the final answer with an example markdown formatting:
Final-answer: ```text
BTWTBIGTKTGBGIKHGTBTBEME
```

Input: P*Y*RC*II*T#*

Output:
解密步骤开始:
1. 接收到的密文: P*Y*RC*II*T#*
2. 根据*号分割密文
  分割得到6个部分
3. 将分割后的内容填入矩阵:
  第1行: P
  第2行: Y
  第3行: RC
  第4行: II
  第5行: T#
  第6行:

4. 交替读取矩阵内容:
  向下读取第1列
  向上读取第2列
  向下读取第3列
  向上读取第4列
  向下读取第5列
  向上读取第6列
  向下读取第7列
  向上读取第8列
  向下读取第9列
  向上读取第10列
  向下读取第11列
  向上读取第12列
  向下读取第13列
5. 移除#号获得最终明文
解密结果: PYRITIC


Prompt:
Please decode the ciphertext according to the examples and the given information.
Please decrypt the password according to the encryption algorithm and decryption algorithm.
Encryption Algorithm:
Encryption Rules:
1. Shift each letter in the input text forward by a specified number of positions (shift).
2. Output the encrypted text.
Decryption Algorithm:
Decryption Rules:
1. Shift each letter in the input text backward by a specified number of positions (shift).
2. Output the decrypted text.
encoded text: qzsjujd secret key or extra parameter: {'shift': 1} decrypt to original information: pyritic
encoded text: qzsjujd secret key or extra parameter: {'shift': 1} decrypt to original information: pyritic
encoded text: qzsjujd secret key or extra parameter: {'shift': 1} decrypt to original information: pyritic
encoded text: ulycpp secret key or extra parameter: {'shift': 7} decrypt to original information: ? Please complete it step by step, ensuring the process is detailed and rigorous, and the result is correct.

Let's think step by step and output the final answer with an example markdown formatting:
Final-answer: ```text
BTWTBIGTKTGBGIKHGTBTBEME
```

Input: ulycpp

Output:
开始解码过程，文本为: ulycpp，位移数为: 7
处理字符: u
字符 u 向前移动 7 位，变为 n
处理字符: l
字符 l 向前移动 7 位，变为 e
处理字符: y
字符 y 向前移动 7 位，变为 r
处理字符: c
字符 c 向前移动 7 位，变为 v
处理字符: p
字符 p 向前移动 7 位，变为 i
处理字符: p
字符 p 向前移动 7 位，变为 i
解码完成，结果为: nervii


Prompt:
Your task is to convert the seemingly meaningless ciphertext into readable information using the appropriate algorithm, ensuring the accuracy and integrity of the information.
Please decrypt the password according to the encryption algorithm and decryption algorithm.
Encryption Algorithm:
Encryption Rules:
1. Shift each letter in the input text forward by a specified number of positions (shift).
2. Output the encrypted text.
Decryption Algorithm:
Decryption Rules:
1. Shift each letter in the input text backward by a specified number of positions (shift).
2. Output the decrypted text.
cipher text: ulycpp secret key or extra parameter: {'shift': 7} decrypt into original information: nervii
cipher text: qzsjujd secret key or extra parameter: {'shift': 1} decrypt into original information: pyritic
cipher text: ulycpp secret key or extra parameter: {'shift': 7} decrypt into original information: nervii
cipher text: cdcrjaepqxaxin secret key or extra parameter: {'shift': 15} decrypt into original information is: ? Please complete it step by step, formulate a reasonable problem-solving plan, and strictly adhere to it.

Let's think step by step and output the final answer with an example markdown formatting:
Final-answer: ```text
BTWTBIGTKTGBGIKHGTBTBEME
```

Input: cdcrjaepqxaxin

Output:
开始解码过程，文本为: cdcrjaepqxaxin，位移数为: 15
处理字符: c
字符 c 向前移动 15 位，变为 n
处理字符: d
字符 d 向前移动 15 位，变为 o
处理字符: c
字符 c 向前移动 15 位，变为 n
处理字符: r
字符 r 向前移动 15 位，变为 c
处理字符: j
字符 j 向前移动 15 位，变为 u
处理字符: a
字符 a 向前移动 15 位，变为 l
处理字符: e
字符 e 向前移动 15 位，变为 p
处理字符: p
字符 p 向前移动 15 位，变为 a
处理字符: q
字符 q 向前移动 15 位，变为 b
处理字符: x
字符 x 向前移动 15 位，变为 i
处理字符: a
字符 a 向前移动 15 位，变为 l
处理字符: x
字符 x 向前移动 15 位，变为 i
处理字符: i
字符 i 向前移动 15 位，变为 t
处理字符: n
字符 n 向前移动 15 位，变为 y
解码完成，结果为: nonculpability


Prompt:
You are an excellent cipher decoder, please refer to the following examples and information to decode the ciphertext.
Please decrypt the password according to the encryption algorithm and decryption algorithm.
Encryption Algorithm:
Encryption Rules:
1. Shift each letter in the input text forward by a specified number of positions (shift).
2. Output the encrypted text.
Decryption Algorithm:
Decryption Rules:
1. Shift each letter in the input text backward by a specified number of positions (shift).
2. Output the decrypted text.
encrypted text: qzsjujd secret key or extra parameter: {'shift': 1} decrypt into plain text: ?  Carefully complete each step.

Let's think step by step and output the final answer with an example markdown formatting:
Final-answer: ```text
BTWTBIGTKTGBGIKHGTBTBEME
```

Input: qzsjujd

Output:
开始解码过程，文本为: qzsjujd，位移数为: 1
处理字符: q
字符 q 向前移动 1 位，变为 p
处理字符: z
字符 z 向前移动 1 位，变为 y
处理字符: s
字符 s 向前移动 1 位，变为 r
处理字符: j
字符 j 向前移动 1 位，变为 i
处理字符: u
字符 u 向前移动 1 位，变为 t
处理字符: j
字符 j 向前移动 1 位，变为 i
处理字符: d
字符 d 向前移动 1 位，变为 c
解码完成，结果为: pyritic


Prompt:
您是一位杰出的密文解密专家，请参考以下案例和信息进行解密操作。
请根据加密算法和解密算法对密码进行解密
加密算法:
加密规则:
- 输入:
    - 原始信息: 大写字母字符串,不含标点和空格
- 输出:
    - 暗码: 大写字母字符串
- 准备:
    - 5x5网格(不含字母J,行列坐标在0到4之间):
        - U B I L A
        N T C D E
        F G H K M
        O P Q R S
        V W X Y Z
- 加密步骤:
    - 去除原始信息中的标点、空格和字母J
    - 对原始信息中的每个字母,找到其在网格中对应的行列坐标(0-4)。例如A是(0,4)
    - 将所有行列坐标重排,先读所有行坐标,再读所有列坐标形成新的坐标序列。例如原来是(0,4)(1,2),现在读作0142
    - 每次从新坐标序列中取出两个数作为新的行列坐标,在网格中找到对应字母形成暗码。例如对于0142,暗码对应(0,1)是B,(4,2)是X。最终暗码是BX
解密算法:
解密规则:
- 输入:
    - 暗码: 大写字母字符串,不含标点和空格
- 输出:
    - 原始信息: 大写字母字符串,不含标点和空格
- 准备:
    - 5x5网格(与加密相同)
- 解密步骤(与加密步骤完全相反):
    - 对暗码中的每个字母,找到其在网格中对应的行列坐标(0-4)得到坐标序列
    - 将坐标序列分成两半,前半部分为所有原始行坐标,后半部分为所有原始列坐标,从原始行坐标读一个,从原始列坐标读一个,根据原始行列坐标在网格中找到对应字母形成原始信息
        - 例如[0,1,4,2],前半部分作为行坐标:[0,1],后半部分作为列坐标:[4,2]
        - 交替读取行坐标和列坐标。得到(0,4),(1,2),分别在网格中找到对应字母为AC
暗码: DTUOUUEUILECQD 解密成为原始信息: nonculpability
暗码: TSUAOH 解密成为原始信息是什么？ 请一步一步完成, 制定合理的解题计划并严格执行。

Let's think step by step and output the final answer with an example markdown formatting:
Final-answer: ```text
BTWTBIGTKTGBGIKHGTBTBEME
```

Input: TSUAOH

Output:
解密步骤:
1. 生成的5x5矩阵:
['U', 'B', 'I', 'L', 'A']
['N', 'T', 'C', 'D', 'E']
['F', 'G', 'H', 'K', 'M']
['O', 'P', 'Q', 'R', 'S']
['V', 'W', 'X', 'Y', 'Z']
2. 获取密文字母的坐标: [1, 3, 0, 0, 3, 2, 1, 4, 0, 4, 0, 2]
3. 还原原始坐标: ['1', '0', '1', '4', '3', '3', '4', '0', '0', '2', '0', '2']
4. 最终解密结果: NERVII


Prompt:
您的主要职责是运用您的专业技能，将提供的密文通过恰当的解密算法恢复成最初的数据形式。
请根据加密算法和解密算法对密码进行解密
加密算法:
加密规则:
- 输入:
    - 原文: 大写字母字符串,不含标点和空格
- 输出:
    - 加密信息: 大写字母字符串
- 准备:
    - 5x5网格(不含字母J,行列坐标在0到4之间):
        - U B I L A
        N T C D E
        F G H K M
        O P Q R S
        V W X Y Z
- 加密步骤:
    - 去除原文中的标点、空格和字母J
    - 对原文中的每个字母,找到其在网格中对应的行列坐标(0-4)。例如A是(0,4)
    - 将所有行列坐标重排,先读所有行坐标,再读所有列坐标形成新的坐标序列。例如原来是(0,4)(1,2),现在读作0142
    - 每次从新坐标序列中取出两个数作为新的行列坐标,在网格中找到对应字母形成加密信息。例如对于0142,加密信息对应(0,1)是B,(4,2)是X。最终加密信息是BX
解密算法:
解密规则:
- 输入:
    - 加密信息: 大写字母字符串,不含标点和空格
- 输出:
    - 原文: 大写字母字符串,不含标点和空格
- 准备:
    - 5x5网格(与加密相同)
- 解密步骤(与加密步骤完全相反):
    - 对加密信息中的每个字母,找到其在网格中对应的行列坐标(0-4)得到坐标序列
    - 将坐标序列分成两半,前半部分为所有原始行坐标,后半部分为所有原始列坐标,从原始行坐标读一个,从原始列坐标读一个,根据原始行列坐标在网格中找到对应字母形成原文
        - 例如[0,1,4,2],前半部分作为行坐标:[0,1],后半部分作为列坐标:[4,2]
        - 交替读取行坐标和列坐标。得到(0,4),(1,2),分别在网格中找到对应字母为AC
加密信息: SONTRGH 解密为原文: pyritic
加密信息: DTUOUUEUILECQD 解密为原文: ? 一步一步完成

Let's think step by step and output the final answer with an example markdown formatting:
Final-answer: ```text
BTWTBIGTKTGBGIKHGTBTBEME
```

Input: DTUOUUEUILECQD

Output:
解密步骤:
1. 生成的5x5矩阵:
['U', 'B', 'I', 'L', 'A']
['N', 'T', 'C', 'D', 'E']
['F', 'G', 'H', 'K', 'M']
['O', 'P', 'Q', 'R', 'S']
['V', 'W', 'X', 'Y', 'Z']
2. 获取密文字母的坐标: [1, 1, 0, 3, 0, 0, 1, 0, 0, 0, 1, 1, 3, 1, 3, 1, 0, 0, 0, 0, 4, 0, 2, 3, 4, 2, 2, 3]
3. 还原原始坐标: ['1', '0', '3', '0', '1', '0', '1', '2', '0', '0', '0', '3', '3', '1', '0', '4', '0', '1', '0', '2', '0', '3', '0', '2', '1', '1', '4', '3']
4. 最终解密结果: NONCULPABILITY


Prompt:
作为解码大师，您的任务是依据案例中描述的解密算法，将密文还原为原始的明文。
请根据加密算法和解密算法对密码进行解密
加密算法:
加密规则:
- 输入:
    - 原文: 大写字母字符串,不含标点和空格
- 输出:
    - 加密文本: 大写字母字符串
- 准备:
    - 5x5网格(不含字母J,行列坐标在0到4之间):
        - U B I L A
        N T C D E
        F G H K M
        O P Q R S
        V W X Y Z
- 加密步骤:
    - 去除原文中的标点、空格和字母J
    - 对原文中的每个字母,找到其在网格中对应的行列坐标(0-4)。例如A是(0,4)
    - 将所有行列坐标重排,先读所有行坐标,再读所有列坐标形成新的坐标序列。例如原来是(0,4)(1,2),现在读作0142
    - 每次从新坐标序列中取出两个数作为新的行列坐标,在网格中找到对应字母形成加密文本。例如对于0142,加密文本对应(0,1)是B,(4,2)是X。最终加密文本是BX
解密算法:
解密规则:
- 输入:
    - 加密文本: 大写字母字符串,不含标点和空格
- 输出:
    - 原文: 大写字母字符串,不含标点和空格
- 准备:
    - 5x5网格(与加密相同)
- 解密步骤(与加密步骤完全相反):
    - 对加密文本中的每个字母,找到其在网格中对应的行列坐标(0-4)得到坐标序列
    - 将坐标序列分成两半,前半部分为所有原始行坐标,后半部分为所有原始列坐标,从原始行坐标读一个,从原始列坐标读一个,根据原始行列坐标在网格中找到对应字母形成原文
        - 例如[0,1,4,2],前半部分作为行坐标:[0,1],后半部分作为列坐标:[4,2]
        - 交替读取行坐标和列坐标。得到(0,4),(1,2),分别在网格中找到对应字母为AC
加密文本: DTUOUUEUILECQD 解码为原文: nonculpability
加密文本: SONTRGH 解码为原文: ?

Let's think step by step and output the final answer with an example markdown formatting:
Final-answer: ```text
BTWTBIGTKTGBGIKHGTBTBEME
```

Input: SONTRGH

Output:
解密步骤:
1. 生成的5x5矩阵:
['U', 'B', 'I', 'L', 'A']
['N', 'T', 'C', 'D', 'E']
['F', 'G', 'H', 'K', 'M']
['O', 'P', 'Q', 'R', 'S']
['V', 'W', 'X', 'Y', 'Z']
2. 获取密文字母的坐标: [3, 3, 1, 1, 3, 2, 2, 4, 0, 0, 1, 3, 1, 2]
3. 还原原始坐标: ['3', '1', '4', '3', '3', '3', '0', '2', '1', '1', '0', '2', '1', '2']
4. 最终解密结果: PYRITIC


Prompt:
Your primary responsibility is to use your professional skills to encode the provided plaintext using the correct algorithm and ensure the security of information during transmission.
**Encryption Rules**

- Input:
    - Plaintext: A string of uppercase letters (A-Z) without punctuation and spaces.
- Output:
    - Ciphertext: A binary string containing only 0s and 1s.
- Preparation:
    - Fixed Key: A binary string of length 8 (e.g., '10101010').
    - Permutation Table:
        - Permutation Table: (2, 0, 3, 1, 4, 6, 5, 7)
        - Inverse Permutation Table: (1, 3, 0, 2, 4, 6, 5, 7)
- Encryption Steps:
    1. Convert Each Character to Binary:
        - For each character in the plaintext, convert it to its ASCII value.
        - Convert this ASCII value to an 8-bit binary string.
    2. XOR Operation:
        - Perform an XOR operation between the 8-bit binary representation of the character and the fixed key.
        - Ensure the key is repeated to match the length of the binary representation if needed.
    3. Permutation:
        - Apply the permutation table to the XOR result to get the final encrypted binary string for each character.
    4. Combine Binary Strings:
        - Concatenate the binary strings of all characters to form the final ciphertext.

original information: pyritic encode to cipher text: 11111100111101011111100011010101111111101101010111011001
original information: nonculpability encode to cipher text: 1101001011010011110100101101100111111111110101101111110011011101110110001101010111010110110101011111111011110101
original information: nervii encode to cipher text is: ?

Let's think step by step and output the final answer with an example markdown formatting:
Final-answer: ```text
BTWTBIGTKTGBGIKHGTBTBEME
```

Input: NERVII

Output:
1. 输入文本处理为: NERVII

处理字符: N
2. 转换为8位二进制: 01001110
3. 与密钥进行XOR运算: 11100100
4. 应用置换表: 11010010

处理字符: E
2. 转换为8位二进制: 01000101
3. 与密钥进行XOR运算: 11101111
4. 应用置换表: 11011111

处理字符: R
2. 转换为8位二进制: 01010010
3. 与密钥进行XOR运算: 11111000
4. 应用置换表: 11111000

处理字符: V
2. 转换为8位二进制: 01010110
3. 与密钥进行XOR运算: 11111100
4. 应用置换表: 11111010

处理字符: I
2. 转换为8位二进制: 01001001
3. 与密钥进行XOR运算: 11100011
4. 应用置换表: 11010101

处理字符: I
2. 转换为8位二进制: 01001001
3. 与密钥进行XOR运算: 11100011
4. 应用置换表: 11010101

最终加密结果: 110100101101111111111000111110101101010111010101


Prompt:
By your deep knowledge in cryptography, your work is to analyze and apply the encryption techniques in the examples, ensuring the security of information during transmission.
**Encryption Rules**

- Input:
    - Plaintext: A string of uppercase letters (A-Z) without punctuation and spaces.
- Output:
    - Ciphertext: A binary string containing only 0s and 1s.
- Preparation:
    - Fixed Key: A binary string of length 8 (e.g., '10101010').
    - Permutation Table:
        - Permutation Table: (2, 0, 3, 1, 4, 6, 5, 7)
        - Inverse Permutation Table: (1, 3, 0, 2, 4, 6, 5, 7)
- Encryption Steps:
    1. Convert Each Character to Binary:
        - For each character in the plaintext, convert it to its ASCII value.
        - Convert this ASCII value to an 8-bit binary string.
    2. XOR Operation:
        - Perform an XOR operation between the 8-bit binary representation of the character and the fixed key.
        - Ensure the key is repeated to match the length of the binary representation if needed.
    3. Permutation:
        - Apply the permutation table to the XOR result to get the final encrypted binary string for each character.
    4. Combine Binary Strings:
        - Concatenate the binary strings of all characters to form the final ciphertext.

clear text: nervii encrypt to encoded text: 110100101101111111111000111110101101010111010101
clear text: nervii encrypt to encoded text: 110100101101111111111000111110101101010111010101
clear text: nervii encrypt to encoded text: 110100101101111111111000111110101101010111010101
clear text: nonculpability encrypt to encoded text is: ?

Let's think step by step and output the final answer with an example markdown formatting:
Final-answer: ```text
BTWTBIGTKTGBGIKHGTBTBEME
```

Input: NONCULPABILITY

Output:
1. 输入文本处理为: NONCULPABILITY

处理字符: N
2. 转换为8位二进制: 01001110
3. 与密钥进行XOR运算: 11100100
4. 应用置换表: 11010010

处理字符: O
2. 转换为8位二进制: 01001111
3. 与密钥进行XOR运算: 11100101
4. 应用置换表: 11010011

处理字符: N
2. 转换为8位二进制: 01001110
3. 与密钥进行XOR运算: 11100100
4. 应用置换表: 11010010

处理字符: C
2. 转换为8位二进制: 01000011
3. 与密钥进行XOR运算: 11101001
4. 应用置换表: 11011001

处理字符: U
2. 转换为8位二进制: 01010101
3. 与密钥进行XOR运算: 11111111
4. 应用置换表: 11111111

处理字符: L
2. 转换为8位二进制: 01001100
3. 与密钥进行XOR运算: 11100110
4. 应用置换表: 11010110

处理字符: P
2. 转换为8位二进制: 01010000
3. 与密钥进行XOR运算: 11111010
4. 应用置换表: 11111100

处理字符: A
2. 转换为8位二进制: 01000001
3. 与密钥进行XOR运算: 11101011
4. 应用置换表: 11011101

处理字符: B
2. 转换为8位二进制: 01000010
3. 与密钥进行XOR运算: 11101000
4. 应用置换表: 11011000

处理字符: I
2. 转换为8位二进制: 01001001
3. 与密钥进行XOR运算: 11100011
4. 应用置换表: 11010101

处理字符: L
2. 转换为8位二进制: 01001100
3. 与密钥进行XOR运算: 11100110
4. 应用置换表: 11010110

处理字符: I
2. 转换为8位二进制: 01001001
3. 与密钥进行XOR运算: 11100011
4. 应用置换表: 11010101

处理字符: T
2. 转换为8位二进制: 01010100
3. 与密钥进行XOR运算: 11111110
4. 应用置换表: 11111110

处理字符: Y
2. 转换为8位二进制: 01011001
3. 与密钥进行XOR运算: 11110011
4. 应用置换表: 11110101

最终加密结果: 1101001011010011110100101101100111111111110101101111110011011101110110001101010111010110110101011111111011110101


Prompt:
As an encoder, your task is to use the encryption algorithm described in the examples to encrypt the plaintext.
**Encryption Rules**

- Input:
    - Plaintext: A string of uppercase letters (A-Z) without punctuation and spaces.
- Output:
    - Ciphertext: A binary string containing only 0s and 1s.
- Preparation:
    - Fixed Key: A binary string of length 8 (e.g., '10101010').
    - Permutation Table:
        - Permutation Table: (2, 0, 3, 1, 4, 6, 5, 7)
        - Inverse Permutation Table: (1, 3, 0, 2, 4, 6, 5, 7)
- Encryption Steps:
    1. Convert Each Character to Binary:
        - For each character in the plaintext, convert it to its ASCII value.
        - Convert this ASCII value to an 8-bit binary string.
    2. XOR Operation:
        - Perform an XOR operation between the 8-bit binary representation of the character and the fixed key.
        - Ensure the key is repeated to match the length of the binary representation if needed.
    3. Permutation:
        - Apply the permutation table to the XOR result to get the final encrypted binary string for each character.
    4. Combine Binary Strings:
        - Concatenate the binary strings of all characters to form the final ciphertext.

plain text: nonculpability encrypt to cipher text: 1101001011010011110100101101100111111111110101101111110011011101110110001101010111010110110101011111111011110101
plain text: nonculpability encrypt to cipher text: 1101001011010011110100101101100111111111110101101111110011011101110110001101010111010110110101011111111011110101
plain text: nonculpability encrypt to cipher text: 1101001011010011110100101101100111111111110101101111110011011101110110001101010111010110110101011111111011110101
plain text: pyritic encrypt to cipher text: ? Please plan the process and implement it step by step, ensuring that each step is carefully checked to ultimately achieve the desired outcome.

Let's think step by step and output the final answer with an example markdown formatting:
Final-answer: ```text
BTWTBIGTKTGBGIKHGTBTBEME
```

Input: PYRITIC

Output:
1. 输入文本处理为: PYRITIC

处理字符: P
2. 转换为8位二进制: 01010000
3. 与密钥进行XOR运算: 11111010
4. 应用置换表: 11111100

处理字符: Y
2. 转换为8位二进制: 01011001
3. 与密钥进行XOR运算: 11110011
4. 应用置换表: 11110101

处理字符: R
2. 转换为8位二进制: 01010010
3. 与密钥进行XOR运算: 11111000
4. 应用置换表: 11111000

处理字符: I
2. 转换为8位二进制: 01001001
3. 与密钥进行XOR运算: 11100011
4. 应用置换表: 11010101

处理字符: T
2. 转换为8位二进制: 01010100
3. 与密钥进行XOR运算: 11111110
4. 应用置换表: 11111110

处理字符: I
2. 转换为8位二进制: 01001001
3. 与密钥进行XOR运算: 11100011
4. 应用置换表: 11010101

处理字符: C
2. 转换为8位二进制: 01000011
3. 与密钥进行XOR运算: 11101001
4. 应用置换表: 11011001

最终加密结果: 11111100111101011111100011010101111111101101010111011001


Prompt:
Please encode the plaintext step by step, ensuring the process is detailed and rigorous, and the result is correct.
Please encrypt the plaintext according to the encryption algorithm.
Encryption rules:
1. Shift each letter in the input text forward by a specified number of positions (shift).
2. Output the encrypted text.
clear text: nervii secret key: {'shift': 7} encrypt into encoded text is: ? Please proceed carefully according to the steps, ensuring that every detail in the process is not overlooked, to achieve an accurate and error-free goal.

Let's think step by step and output the final answer with an example markdown formatting:
Final-answer: ```text
BTWTBIGTKTGBGIKHGTBTBEME
```

Input: nervii

Output:
开始编码过程，文本为: nervii，位移数为: 7
处理字符: n
字符 n 向后移动 7 位，变为 u
处理字符: e
字符 e 向后移动 7 位，变为 l
处理字符: r
字符 r 向后移动 7 位，变为 y
处理字符: v
字符 v 向后移动 7 位，变为 c
处理字符: i
字符 i 向后移动 7 位，变为 p
处理字符: i
字符 i 向后移动 7 位，变为 p
编码完成，结果为: ulycpp


Prompt:
Please encode the plaintext step by step, ensuring the process is detailed and rigorous, and the result is correct.
Please encrypt the plaintext according to the encryption algorithm.
Encryption rules:
1. Shift each letter in the input text forward by a specified number of positions (shift).
2. Output the encrypted text.
original information: nervii secret key: {'shift': 7} encrypt into encoded text: ulycpp
original information: nervii secret key: {'shift': 7} encrypt into encoded text: ulycpp
original information: nonculpability secret key: {'shift': 15} encrypt into encoded text: ?  Carefully complete each step.

Let's think step by step and output the final answer with an example markdown formatting:
Final-answer: ```text
BTWTBIGTKTGBGIKHGTBTBEME
```

Input: nonculpability

Output:
开始编码过程，文本为: nonculpability，位移数为: 15
处理字符: n
字符 n 向后移动 15 位，变为 c
处理字符: o
字符 o 向后移动 15 位，变为 d
处理字符: n
字符 n 向后移动 15 位，变为 c
处理字符: c
字符 c 向后移动 15 位，变为 r
处理字符: u
字符 u 向后移动 15 位，变为 j
处理字符: l
字符 l 向后移动 15 位，变为 a
处理字符: p
字符 p 向后移动 15 位，变为 e
处理字符: a
字符 a 向后移动 15 位，变为 p
处理字符: b
字符 b 向后移动 15 位，变为 q
处理字符: i
字符 i 向后移动 15 位，变为 x
处理字符: l
字符 l 向后移动 15 位，变为 a
处理字符: i
字符 i 向后移动 15 位，变为 x
处理字符: t
字符 t 向后移动 15 位，变为 i
处理字符: y
字符 y 向后移动 15 位，变为 n
编码完成，结果为: cdcrjaepqxaxin


Prompt:
As an encoder, your task is to use the encryption algorithm described in the examples to encrypt the plaintext.
Please encrypt the plaintext according to the encryption algorithm.
Encryption rules:
1. Shift each letter in the input text forward by a specified number of positions (shift).
2. Output the encrypted text.
clear text: nonculpability extra parameter: {'shift': 15} encode to encoded text: cdcrjaepqxaxin
clear text: nonculpability extra parameter: {'shift': 15} encode to encoded text: cdcrjaepqxaxin
clear text: pyritic extra parameter: {'shift': 1} encode to encoded text: ? Please complete it step by step, formulate a reasonable problem-solving plan, and strictly adhere to it.

Let's think step by step and output the final answer with an example markdown formatting:
Final-answer: ```text
BTWTBIGTKTGBGIKHGTBTBEME
```

Input: pyritic

Output:
开始编码过程，文本为: pyritic，位移数为: 1
处理字符: p
字符 p 向后移动 1 位，变为 q
处理字符: y
字符 y 向后移动 1 位，变为 z
处理字符: r
字符 r 向后移动 1 位，变为 s
处理字符: i
字符 i 向后移动 1 位，变为 j
处理字符: t
字符 t 向后移动 1 位，变为 u
处理字符: i
字符 i 向后移动 1 位，变为 j
处理字符: c
字符 c 向后移动 1 位，变为 d
编码完成，结果为: qzsjujd


Prompt:
作为解码大师，您的任务是依据案例中描述的解密算法，将密文还原为原始的明文。
请根据加密算法和解密算法对密码进行解密
加密算法:
加密规则:
- 输入:
    - 明文: 不含标点和空格的大写字母字符串
- 输出:
    - 密文: 由逗号分隔的数字序列，例如"y1,y2,..."
- 准备:
    - k_q_x值: 12
- 加密步骤:
    - 对明文中的每个字母p:
        - 获取字母p对应的ASCII码值x
        - 计算x * k_q_x作为该字母对应的密文数字y
    - 最后将所有y值用逗号连接得到最终密文
解密算法:
解密规则:
- 输入:
    - 密文: 由逗号分隔的数字序列，例如"y1,y2,..."
- 输出:
    - 明文: 不含标点和空格的大写字母字符串
- 准备:
    - k_q_x值: 12
- 解密步骤:
    - 对密文中的每个数字c:
        - 计算z = c // k_q_x，其中//表示整数除法运算，返回商的整数部分
        - 根据z对应的ASCII码值找到对应的字母作为明文字母p
    - 最后将所有p连接得到最终明文
密文: 960,1068,984,876,1008,876,804 解密成为明文: pyritic
密文: 936,948,936,804,1020,912,960,780,792,876,912,876,1008,1068 解密成为明文: nonculpability
密文: 960,1068,984,876,1008,876,804 解密成为明文: pyritic
密文: 936,828,984,1032,876,876 解密成为明文是什么？ 精心完成每一步。

Let's think step by step and output the final answer with an example markdown formatting:
Final-answer: ```text
BTWTBIGTKTGBGIKHGTBTBEME
```

Output:
开始解密过程...
收到的加密文本: 936,828,984,1032,876,876
使用预设的k_q_x值: 12

处理数字: 936
除以k_q_x后的值: 78
对应的字符: N

处理数字: 828
除以k_q_x后的值: 69
对应的字符: E

处理数字: 984
除以k_q_x后的值: 82
对应的字符: R

处理数字: 1032
除以k_q_x后的值: 86
对应的字符: V

处理数字: 876
除以k_q_x后的值: 73
对应的字符: I

处理数字: 876
除以k_q_x后的值: 73
对应的字符: I

最终解密结果: NERVII


Prompt:
请根据加密算法和解密算法对密码进行解密
加密算法:
加密规则:
- 输入:
    - 原始信息: 不含标点和空格的大写字母字符串
- 输出:
    - 暗码: 由逗号分隔的数字序列，例如"y1,y2,..."
- 准备:
    - k_q_x值: 12
- 加密步骤:
    - 对原始信息中的每个字母p:
        - 获取字母p对应的ASCII码值x
        - 计算x * k_q_x作为该字母对应的暗码数字y
    - 最后将所有y值用逗号连接得到最终暗码
解密算法:
解密规则:
- 输入:
    - 暗码: 由逗号分隔的数字序列，例如"y1,y2,..."
- 输出:
    - 原始信息: 不含标点和空格的大写字母字符串
- 准备:
    - k_q_x值: 12
- 解密步骤:
    - 对暗码中的每个数字c:
        - 计算z = c // k_q_x，其中//表示整数除法运算，返回商的整数部分
        - 根据z对应的ASCII码值找到对应的字母作为原始信息字母p
    - 最后将所有p连接得到最终原始信息
暗码: 936,948,936,804,1020,912,960,780,792,876,912,876,1008,1068 解密成原始信息: ? 精心完成每一步。

Let's think step by step and output the final answer with an example markdown formatting:
Final-answer: ```text
BTWTBIGTKTGBGIKHGTBTBEME
```

Input: 936,948,936,804,1020,912,960,780,792,876,912,876,1008,1068

Output:
开始解密过程...
收到的加密文本: 936,948,936,804,1020,912,960,780,792,876,912,876,1008,1068
使用预设的k_q_x值: 12

处理数字: 936
除以k_q_x后的值: 78
对应的字符: N

处理数字: 948
除以k_q_x后的值: 79
对应的字符: O

处理数字: 936
除以k_q_x后的值: 78
对应的字符: N

处理数字: 804
除以k_q_x后的值: 67
对应的字符: C

处理数字: 1020
除以k_q_x后的值: 85
对应的字符: U

处理数字: 912
除以k_q_x后的值: 76
对应的字符: L

处理数字: 960
除以k_q_x后的值: 80
对应的字符: P

处理数字: 780
除以k_q_x后的值: 65
对应的字符: A

处理数字: 792
除以k_q_x后的值: 66
对应的字符: B

处理数字: 876
除以k_q_x后的值: 73
对应的字符: I

处理数字: 912
除以k_q_x后的值: 76
对应的字符: L

处理数字: 876
除以k_q_x后的值: 73
对应的字符: I

处理数字: 1008
除以k_q_x后的值: 84
对应的字符: T

处理数字: 1068
除以k_q_x后的值: 89
对应的字符: Y

最终解密结果: NONCULPABILITY


Prompt:
您是一位杰出的密文解密专家，请参考以下案例和信息进行解密操作。
请根据加密算法和解密算法对密码进行解密
加密算法:
加密规则:
- 输入:
    - 非加密信息: 不含标点和空格的大写字母字符串
- 输出:
    - 暗码: 由逗号分隔的数字序列，例如"y1,y2,..."
- 准备:
    - k_q_x值: 12
- 加密步骤:
    - 对非加密信息中的每个字母p:
        - 获取字母p对应的ASCII码值x
        - 计算x * k_q_x作为该字母对应的暗码数字y
    - 最后将所有y值用逗号连接得到最终暗码
解密算法:
解密规则:
- 输入:
    - 暗码: 由逗号分隔的数字序列，例如"y1,y2,..."
- 输出:
    - 非加密信息: 不含标点和空格的大写字母字符串
- 准备:
    - k_q_x值: 12
- 解密步骤:
    - 对暗码中的每个数字c:
        - 计算z = c // k_q_x，其中//表示整数除法运算，返回商的整数部分
        - 根据z对应的ASCII码值找到对应的字母作为非加密信息字母p
    - 最后将所有p连接得到最终非加密信息
暗码: 936,828,984,1032,876,876 解密成为非加密信息: nervii
暗码: 936,948,936,804,1020,912,960,780,792,876,912,876,1008,1068 解密成为非加密信息: nonculpability
暗码: 960,1068,984,876,1008,876,804 解密成为非加密信息: ? 精心完成每一步。

Let's think step by step and output the final answer with an example markdown formatting:
Final-answer: ```text
BTWTBIGTKTGBGIKHGTBTBEME
```

Output:
开始解密过程...
收到的加密文本: 960,1068,984,876,1008,876,804
使用预设的k_q_x值: 12

处理数字: 960
除以k_q_x后的值: 80
对应的字符: P

处理数字: 1068
除以k_q_x后的值: 89
对应的字符: Y

处理数字: 984
除以k_q_x后的值: 82
对应的字符: R

处理数字: 876
除以k_q_x后的值: 73
对应的字符: I

处理数字: 1008
除以k_q_x后的值: 84
对应的字符: T

处理数字: 876
除以k_q_x后的值: 73
对应的字符: I

处理数字: 804
除以k_q_x后的值: 67
对应的字符: C

最终解密结果: PYRITIC


Prompt:
**Encryption Rules:**

- Input:
    - Plaintext: A string of uppercase letters and spaces.
- Output:
    - Ciphertext: A hexadecimal string representing the encrypted data (where A-E needs to be capitalised).
- Preparation:
    - Fixed key
        - KEY = b'1234567890ABCDEF'
    - S_BOX
        - S_BOX = {
        0x00: 0x0F, 0x01: 0x0A, 0x02: 0x07, 0x03: 0x05,
        0x04: 0x09, 0x05: 0x03, 0x06: 0x0D, 0x07: 0x00,
        0x08: 0x0E, 0x09: 0x08, 0x0A: 0x04, 0x0B: 0x06,
        0x0C: 0x01, 0x0D: 0x02, 0x0E: 0x0B, 0x0F: 0x0C
        }
- Encryption Steps:
    1. Padding: If the plaintext is not a multiple of 8 bytes, pad it with \x00 (null character) to make its length a multiple of 8 bytes.
    2. Block Splitting: Divide the padded plaintext into 8-byte blocks.
    3. Block Encryption:
        - Convert Block to Bytes: Convert each block to bytes using ASCII encoding.
        - XOR with Key: Apply XOR between the byte block and the fixed key.
        - Substitute:  Use the S_BOX Replacement Box to replace the high 4 bits and low 4 bits in each byte before stitching them together.
        - Permute: Perform a simple permutation by rotating each byte left by 1 bit.
        - XOR with Key: Apply XOR between the permuted byte block and the fixed key again.
    4. Hexadecimal Encoding: Convert the encrypted byte blocks to a hexadecimal string.
    5. Concatenate: Concatenate all the hexadecimal strings of the encrypted blocks to form the final ciphertext.

**Decryption Rules:**

- Input:
    - Ciphertext: A hexadecimal string representing the encrypted data (where A-E needs to be capitalised).
- Output:
    - Plaintext: A string of uppercase letters and spaces.
- Preparation:
    - Fixed key
        - KEY = b'1234567890ABCDEF'
    - INV_S_BOX
        - INV_S_BOX = {
        0x0F: 0x00, 0x0A: 0x01, 0x07: 0x02, 0x05: 0x03,
        0x09: 0x04, 0x03: 0x05, 0x0D: 0x06, 0x00: 0x07,
        0x0E: 0x08, 0x08: 0x09, 0x04: 0x0A, 0x06: 0x0B,
        0x01: 0x0C, 0x02: 0x0D, 0x0B: 0x0E, 0x0C: 0x0F
        }
- Decryption Steps:
    1. Block Splitting: Divide the ciphertext into 16-character (8-byte) blocks (each block is 8 bytes long).
    2. Block Decryption:
        - Convert Hex to Bytes: Convert each block from hexadecimal string to bytes.
        - XOR with Key: Apply XOR between the byte block and the fixed key.
        - Inverse Permute: Perform an inverse permutation by rotating each byte right by 1 bit.
        - Substitute: Use INV_S_BOX to replace the high four and low four bits in a block of bytes and splice them together.
        - XOR with Key: Apply XOR between the byte block and the fixed key again.
    3. Convert Bytes to Text: Convert the decrypted byte blocks back to text using ASCII decoding.
    4. Remove Padding: Remove any padding characters (\x00) from the end of the decrypted plaintext.
    5. Concatenate: Concatenate all the decrypted blocks to form the final plaintext.
encoded text: 293637348A3E96283B3E2B30808F9784 decrypt to plain text: nonculpability
encoded text: 293637348A3E96283B3E2B30808F9784 decrypt to plain text: nonculpability
encoded text: 2932869B372E9784 decrypt to plain text is: ? step by step

Let's think step by step and output the final answer with an example markdown formatting:
Final-answer: ```text
BTWTBIGTKTGBGIKHGTBTBEME
```

Output:
开始解密过程...
1. 接收到的密文: 2932869B372E9784
2. 开始分块处理...

处理第1个块: 2932869B372E9784
  - 转换为字节: 2932869B372E9784
  - 与密钥XOR后: 1800B5AF0218A0BC
  - 右移置换后: 0C00DAD7010C505E
  - 逆S盒替换后: 7F7761627C7F3738
  - 最终与密钥XOR后: 4E45525649490000

最终解密结果: NERVII


Prompt:
As a specialist in cryptography, your job is to analyze the encryption algorithm in the examples and implement the decryption process on the data.
**Encryption Rules:**

- Input:
    - Plaintext: A string of uppercase letters and spaces.
- Output:
    - Ciphertext: A hexadecimal string representing the encrypted data (where A-E needs to be capitalised).
- Preparation:
    - Fixed key
        - KEY = b'1234567890ABCDEF'
    - S_BOX
        - S_BOX = {
        0x00: 0x0F, 0x01: 0x0A, 0x02: 0x07, 0x03: 0x05,
        0x04: 0x09, 0x05: 0x03, 0x06: 0x0D, 0x07: 0x00,
        0x08: 0x0E, 0x09: 0x08, 0x0A: 0x04, 0x0B: 0x06,
        0x0C: 0x01, 0x0D: 0x02, 0x0E: 0x0B, 0x0F: 0x0C
        }
- Encryption Steps:
    1. Padding: If the plaintext is not a multiple of 8 bytes, pad it with \x00 (null character) to make its length a multiple of 8 bytes.
    2. Block Splitting: Divide the padded plaintext into 8-byte blocks.
    3. Block Encryption:
        - Convert Block to Bytes: Convert each block to bytes using ASCII encoding.
        - XOR with Key: Apply XOR between the byte block and the fixed key.
        - Substitute:  Use the S_BOX Replacement Box to replace the high 4 bits and low 4 bits in each byte before stitching them together.
        - Permute: Perform a simple permutation by rotating each byte left by 1 bit.
        - XOR with Key: Apply XOR between the permuted byte block and the fixed key again.
    4. Hexadecimal Encoding: Convert the encrypted byte blocks to a hexadecimal string.
    5. Concatenate: Concatenate all the hexadecimal strings of the encrypted blocks to form the final ciphertext.

**Decryption Rules:**

- Input:
    - Ciphertext: A hexadecimal string representing the encrypted data (where A-E needs to be capitalised).
- Output:
    - Plaintext: A string of uppercase letters and spaces.
- Preparation:
    - Fixed key
        - KEY = b'1234567890ABCDEF'
    - INV_S_BOX
        - INV_S_BOX = {
        0x0F: 0x00, 0x0A: 0x01, 0x07: 0x02, 0x05: 0x03,
        0x09: 0x04, 0x03: 0x05, 0x0D: 0x06, 0x00: 0x07,
        0x0E: 0x08, 0x08: 0x09, 0x04: 0x0A, 0x06: 0x0B,
        0x01: 0x0C, 0x02: 0x0D, 0x0B: 0x0E, 0x0C: 0x0F
        }
- Decryption Steps:
    1. Block Splitting: Divide the ciphertext into 16-character (8-byte) blocks (each block is 8 bytes long).
    2. Block Decryption:
        - Convert Hex to Bytes: Convert each block from hexadecimal string to bytes.
        - XOR with Key: Apply XOR between the byte block and the fixed key.
        - Inverse Permute: Perform an inverse permutation by rotating each byte right by 1 bit.
        - Substitute: Use INV_S_BOX to replace the high four and low four bits in a block of bytes and splice them together.
        - XOR with Key: Apply XOR between the byte block and the fixed key again.
    3. Convert Bytes to Text: Convert the decrypted byte blocks back to text using ASCII decoding.
    4. Remove Padding: Remove any padding characters (\x00) from the end of the decrypted plaintext.
    5. Concatenate: Concatenate all the decrypted blocks to form the final plaintext.
encoded text: 2932869B372E9784 decode to clear text: nervii
encoded text: 2932869B372E9784 decode to clear text: nervii
encoded text: 849F8630802E2584 decode to clear text: pyritic
encoded text: 293637348A3E96283B3E2B30808F9784 decode to clear text: ?

Let's think step by step and output the final answer with an example markdown formatting:
Final-answer: ```text
BTWTBIGTKTGBGIKHGTBTBEME
```

Output:
开始解密过程...
1. 接收到的密文: 293637348A3E96283B3E2B30808F9784
2. 开始分块处理...

处理第1个块: 293637348A3E9628
  - 转换为字节: 293637348A3E9628
  - 与密钥XOR后: 18040400BF08A110
  - 右移置换后: 0C020200DF04D008
  - 逆S盒替换后: 7F7D7D77607A6779
  - 最终与密钥XOR后: 4E4F4E43554C5041

处理第2个块: 3B3E2B30808F9784
  - 转换为字节: 3B3E2B30808F9784
  - 与密钥XOR后: 0A0C1804B5B9A0BC
  - 右移置换后: 05060C02DADC505E
  - 逆S盒替换后: 737B7F7D616F3738
  - 最终与密钥XOR后: 42494C4954590000

最终解密结果: NONCULPABILITY


Prompt:
By your deep knowledge in cryptography, your work is to analyze and apply the encryption techniques in the examples, ensuring the security of information during transmission.
**Encryption Rules:**

- Input:
    - Plaintext: A string of uppercase letters and spaces.
- Output:
    - Ciphertext: A hexadecimal string representing the encrypted data (where A-E needs to be capitalised).
- Preparation:
    - Fixed key
        - KEY = b'1234567890ABCDEF'
    - S_BOX
        - S_BOX = {
        0x00: 0x0F, 0x01: 0x0A, 0x02: 0x07, 0x03: 0x05,
        0x04: 0x09, 0x05: 0x03, 0x06: 0x0D, 0x07: 0x00,
        0x08: 0x0E, 0x09: 0x08, 0x0A: 0x04, 0x0B: 0x06,
        0x0C: 0x01, 0x0D: 0x02, 0x0E: 0x0B, 0x0F: 0x0C
        }
- Encryption Steps:
    1. Padding: If the plaintext is not a multiple of 8 bytes, pad it with \x00 (null character) to make its length a multiple of 8 bytes.
    2. Block Splitting: Divide the padded plaintext into 8-byte blocks.
    3. Block Encryption:
        - Convert Block to Bytes: Convert each block to bytes using ASCII encoding.
        - XOR with Key: Apply XOR between the byte block and the fixed key.
        - Substitute:  Use the S_BOX Replacement Box to replace the high 4 bits and low 4 bits in each byte before stitching them together.
        - Permute: Perform a simple permutation by rotating each byte left by 1 bit.
        - XOR with Key: Apply XOR between the permuted byte block and the fixed key again.
    4. Hexadecimal Encoding: Convert the encrypted byte blocks to a hexadecimal string.
    5. Concatenate: Concatenate all the hexadecimal strings of the encrypted blocks to form the final ciphertext.

**Decryption Rules:**

- Input:
    - Ciphertext: A hexadecimal string representing the encrypted data (where A-E needs to be capitalised).
- Output:
    - Plaintext: A string of uppercase letters and spaces.
- Preparation:
    - Fixed key
        - KEY = b'1234567890ABCDEF'
    - INV_S_BOX
        - INV_S_BOX = {
        0x0F: 0x00, 0x0A: 0x01, 0x07: 0x02, 0x05: 0x03,
        0x09: 0x04, 0x03: 0x05, 0x0D: 0x06, 0x00: 0x07,
        0x0E: 0x08, 0x08: 0x09, 0x04: 0x0A, 0x06: 0x0B,
        0x01: 0x0C, 0x02: 0x0D, 0x0B: 0x0E, 0x0C: 0x0F
        }
- Decryption Steps:
    1. Block Splitting: Divide the ciphertext into 16-character (8-byte) blocks (each block is 8 bytes long).
    2. Block Decryption:
        - Convert Hex to Bytes: Convert each block from hexadecimal string to bytes.
        - XOR with Key: Apply XOR between the byte block and the fixed key.
        - Inverse Permute: Perform an inverse permutation by rotating each byte right by 1 bit.
        - Substitute: Use INV_S_BOX to replace the high four and low four bits in a block of bytes and splice them together.
        - XOR with Key: Apply XOR between the byte block and the fixed key again.
    3. Convert Bytes to Text: Convert the decrypted byte blocks back to text using ASCII decoding.
    4. Remove Padding: Remove any padding characters (\x00) from the end of the decrypted plaintext.
    5. Concatenate: Concatenate all the decrypted blocks to form the final plaintext.
encoded text: 2932869B372E9784 decode to clear text: nervii
encoded text: 293637348A3E96283B3E2B30808F9784 decode to clear text: nonculpability
encoded text: 849F8630802E2584 decode to clear text: ?  Carefully complete each step.

Let's think step by step and output the final answer with an example markdown formatting:
Final-answer: ```text
BTWTBIGTKTGBGIKHGTBTBEME
```

Output:
开始解密过程...
1. 接收到的密文: 849F8630802E2584
2. 开始分块处理...

处理第1个块: 849F8630802E2584
  - 转换为字节: 849F8630802E2584
  - 与密钥XOR后: B5ADB504B51812BC
  - 右移置换后: DAD6DA02DA0C095E
  - 逆S盒替换后: 616B617D617F7438
  - 最终与密钥XOR后: 5059524954494300

最终解密结果: PYRITIC


Prompt:
您的任务是使用正确的算法将看似无意义的密文转换回可读的原始信息，确保信息的准确性和完整性。
请根据加密算法和解密算法对密码进行解密
加密算法:
加密规则:
- 输入:
    - 原文: 大写字母字符串，不含标点和空格
- 输出:
    - 隐文: 不含标点和空格的字符串
- 准备:
    - 每行最大字符数: 5
- 加密步骤:
    - 行数从1开始计数
    - 原文按特殊方式排列：奇数行从左到右写，偶数行从右到左写，每行最多五个字母
        - 例如，对于原文"LIDAHELLOWORLD"，先从左到右写第一行为LIDAH，然后从右到左写第二行为WOLLE，然后从左到右写第三行为ORLD，写完的全部内容表示如下
        LIDAH
        WOLLE
        ORLD
    - 然后按列读取，每列从上到下读取，每读完一列加一个"#"，读出的内容即为最终隐文
        - 例如，上述写好的内容按列读取为LWO#IOR#DLL#ALD#HE#
解密算法:
解密规则:
- 输入:
    - 隐文: 不含标点和空格的字符串
- 输出:
    - 原文: 大写字母字符串，不含标点和空格
- 准备:
    - 行数: 5
- 解密步骤(与加密步骤完全相反):
    - 对隐文中的每个字符，从上到下写入；如果遇到#，则切换到下一列继续写入，直到整个隐文写完
        - 例如，对于隐文LWO#IOR#DLL#ALD#HE#，写出如下
        LIDAH
        WOLLE
        ORLD
    - 然后按照奇数行从左到右读，偶数行从右到左读的顺序依次读取，最终结果即为解密后的原文
        - 例如，对于上述写好的内容，第一行从左到右读为LIDAH，第二行从右到左读为ELLOW，第三行从左到右读为ORLD，最终全部内容连接起来，解密原文为LIDAHELLOWORLD
隐文: N#E#R#V#II# 解密成为原文: ? 一步一步完成

Let's think step by step and output the final answer with an example markdown formatting:
Final-answer: ```text
BTWTBIGTKTGBGIKHGTBTBEME
```

Input: N#E#R#V#II#

Output:
1. 收到密文: N#E#R#V#II#
2. 将密文分割成5列: ['N', 'E', 'R', 'V', 'II']
3. 按列填充网格:
   第1列: N
   第2列: E
   第3列: R
   第4列: V
   第5列: I I
4. 按特定顺序读取:
   第1行从左到右读取: N E R V I
   第2行从右到左读取: I
5. 最终明文: NERVII