Prompt:
You are an excellent cipher decoder, please refer to the following examples and information to decode the ciphertext.
**Encryption Rules:**

- Input:
    - Plaintext: Uppercase letters string without punctuation and spaces.
- Output:
    - Ciphertext: A string without punctuation.
- Preparation:
    - Multitap Code Table
        | Letter | Multitap Code |
        | --- | --- |
        | A | 2^1 |
        | B | 2^2 |
        | C | 2^3 |
        | D | 3^1 |
        | E | 3^2 |
        | F | 3^3 |
        | G | 4^1 |
        | H | 4^2 |
        | I | 4^3 |
        | J | 5^1 |
        | K | 5^2 |
        | L | 5^3 |
        | M | 6^1 |
        | N | 6^2 |
        | O | 6^3 |
        | P | 7^1 |
        | Q | 7^2 |
        | R | 7^3 |
        | S | 7^4 |
        | T | 8^1 |
        | U | 8^2 |
        | V | 8^3 |
        | W | 9^1 |
        | X | 9^2 |
        | Y | 9^3 |
        | Z | 9^4 |
- Encryption Steps:
    - For each given plaintext character p:
        - If `p` is an uppercase letter and exists in the Multitap Code Table:
            - Replace `p` with the corresponding Multitap Code from the Multitap Code Table.

 **Decryption Rules:**

- Input:
    - Ciphertext: A string without punctuation.
- Output:
    - Plaintext: Uppercase letters string.
- Preparation: Multitap Code Table (Same as encryption)
- Decryption Steps (exact opposite of encryption steps):
    - For each given ciphertext Multitap Code c:
        - If `c` is a Multitap Code from the Multitap Code Table:
            - Replace `c` with the corresponding uppercase letter from the Multitap Code Table.
encoded text: 7^19^37^34^38^14^32^3 decode into original information: pyritic
encoded text: 6^23^27^38^34^34^3 decode into original information: nervii
encoded text: 6^26^36^22^38^25^37^12^12^24^35^34^38^19^3 decode into original information: ? 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: 6^26^36^22^38^25^37^12^12^24^35^34^38^19^3

Output:
加密文本: 6^26^36^22^38^25^37^12^12^24^35^34^38^19^3
开始解密:
多击编码 6^2 对应的字符是: N
多击编码 6^3 对应的字符是: O
多击编码 6^2 对应的字符是: N
多击编码 2^3 对应的字符是: C
多击编码 8^2 对应的字符是: U
多击编码 5^3 对应的字符是: L
多击编码 7^1 对应的字符是: P
多击编码 2^1 对应的字符是: A
多击编码 2^2 对应的字符是: B
多击编码 4^3 对应的字符是: I
多击编码 5^3 对应的字符是: L
多击编码 4^3 对应的字符是: I
多击编码 8^1 对应的字符是: T
多击编码 9^3 对应的字符是: Y
解密结果: NONCULPABILITY


Prompt:
You are an excellent cipher decoder, please refer to the following examples and information to decode the ciphertext.
**Encryption Rules:**

- Input:
    - Plaintext: Uppercase letters string without punctuation and spaces.
- Output:
    - Ciphertext: A string without punctuation.
- Preparation:
    - Multitap Code Table
        | Letter | Multitap Code |
        | --- | --- |
        | A | 2^1 |
        | B | 2^2 |
        | C | 2^3 |
        | D | 3^1 |
        | E | 3^2 |
        | F | 3^3 |
        | G | 4^1 |
        | H | 4^2 |
        | I | 4^3 |
        | J | 5^1 |
        | K | 5^2 |
        | L | 5^3 |
        | M | 6^1 |
        | N | 6^2 |
        | O | 6^3 |
        | P | 7^1 |
        | Q | 7^2 |
        | R | 7^3 |
        | S | 7^4 |
        | T | 8^1 |
        | U | 8^2 |
        | V | 8^3 |
        | W | 9^1 |
        | X | 9^2 |
        | Y | 9^3 |
        | Z | 9^4 |
- Encryption Steps:
    - For each given plaintext character p:
        - If `p` is an uppercase letter and exists in the Multitap Code Table:
            - Replace `p` with the corresponding Multitap Code from the Multitap Code Table.

 **Decryption Rules:**

- Input:
    - Ciphertext: A string without punctuation.
- Output:
    - Plaintext: Uppercase letters string.
- Preparation: Multitap Code Table (Same as encryption)
- Decryption Steps (exact opposite of encryption steps):
    - For each given ciphertext Multitap Code c:
        - If `c` is a Multitap Code from the Multitap Code Table:
            - Replace `c` with the corresponding uppercase letter from the Multitap Code Table.
cipher text: 6^26^36^22^38^25^37^12^12^24^35^34^38^19^3 decode into original information: nonculpability
cipher text: 6^26^36^22^38^25^37^12^12^24^35^34^38^19^3 decode into original information: nonculpability
cipher text: 7^19^37^34^38^14^32^3 decode into original information: ? 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
```

Output:
加密文本: 7^19^37^34^38^14^32^3
开始解密:
多击编码 7^1 对应的字符是: P
多击编码 9^3 对应的字符是: Y
多击编码 7^3 对应的字符是: R
多击编码 4^3 对应的字符是: I
多击编码 8^1 对应的字符是: T
多击编码 4^3 对应的字符是: I
多击编码 2^3 对应的字符是: C
解密结果: PYRITIC


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:
    - 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.

original information: nonculpability encode to encrypted text: FMMLWKFPPPDEOF
original information: nonculpability encode to encrypted text: FMMLWKFPPPDEOF
original information: nervii encode to encrypted 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: NERVII

Output:
处理后的输入文本: NERVII
开始加密过程:

处理第1个字符 'N':
字符'N'在字母表中的位置是: 23
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
加密计算: (23 + 14) % 26 = 11
加密后的字符是: F

处理第2个字符 'E':
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: 7
加密计算: (13 + 7) % 26 = 20
加密后的字符是: B

处理第3个字符 'R':
字符'R'在字母表中的位置是: 5
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
加密计算: (5 + 13) % 26 = 18
加密后的字符是: P

处理第4个字符 'V':
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: 17
加密计算: (19 + 17) % 26 = 10
加密后的字符是: M

处理第5个字符 'I':
字符'I'在字母表中的位置是: 25
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
加密计算: (25 + 38) % 26 = 11
加密后的字符是: F

处理第6个字符 'I':
字符'I'在字母表中的位置是: 25
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
加密计算: (25 + 45) % 26 = 18
加密后的字符是: P

最终加密结果: FBPMFP


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: 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.

original information: nervii encode into encrypted text: FBPMFP
original information: nonculpability encode into encrypted 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: NONCULPABILITY

Output:
处理后的输入文本: NONCULPABILITY
开始加密过程:

处理第1个字符 'N':
字符'N'在字母表中的位置是: 23
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
加密计算: (23 + 14) % 26 = 11
加密后的字符是: F

处理第2个字符 'O':
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: 7
加密计算: (3 + 7) % 26 = 10
加密后的字符是: M

处理第3个字符 'N':
字符'N'在字母表中的位置是: 23
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
加密计算: (23 + 13) % 26 = 10
加密后的字符是: M

处理第4个字符 'C':
字符'C'在字母表中的位置是: 7
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
加密计算: (7 + 17) % 26 = 24
加密后的字符是: L

处理第5个字符 'U':
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
加密计算: (16 + 38) % 26 = 2
加密后的字符是: W

处理第6个字符 'L':
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: 45
加密计算: (24 + 45) % 26 = 17
加密后的字符是: K

处理第7个字符 'P':
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
加密计算: (18 + 45) % 26 = 11
加密后的字符是: F

处理第8个字符 'A':
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: 38
加密计算: (6 + 38) % 26 = 18
加密后的字符是: P

处理第9个字符 'B':
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: 50
加密计算: (20 + 50) % 26 = 18
加密后的字符是: P

处理第10个字符 'I':
字符'I'在字母表中的位置是: 25
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
加密计算: (25 + 45) % 26 = 18
加密后的字符是: P

处理第11个字符 'L':
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: 29
加密计算: (24 + 29) % 26 = 1
加密后的字符是: D

处理第12个字符 'I':
字符'I'在字母表中的位置是: 25
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
加密计算: (25 + 14) % 26 = 13
加密后的字符是: E

处理第13个字符 'T':
字符'T'在字母表中的位置是: 4
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
加密计算: (4 + 25) % 26 = 3
加密后的字符是: O

处理第14个字符 'Y':
字符'Y'在字母表中的位置是: 12
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
加密计算: (12 + 25) % 26 = 11
加密后的字符是: F

最终加密结果: FMMLWKFPPPDEOF


Prompt:
Please encode the plaintext step by step, ensuring the process is detailed and rigorous, and the result is correct.
**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.

clear text: nervii encrypt into encrypted text: FBPMFP
clear text: nervii encrypt into encrypted text: FBPMFP
clear text: nonculpability encrypt into encrypted text: FMMLWKFPPPDEOF
clear text: pyritic encrypt into encrypted 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: PYRITIC

Output:
处理后的输入文本: PYRITIC
开始加密过程:

处理第1个字符 'P':
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: 14
加密计算: (18 + 14) % 26 = 6
加密后的字符是: A

处理第2个字符 'Y':
字符'Y'在字母表中的位置是: 12
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
加密计算: (12 + 7) % 26 = 19
加密后的字符是: V

处理第3个字符 'R':
字符'R'在字母表中的位置是: 5
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
加密计算: (5 + 13) % 26 = 18
加密后的字符是: P

处理第4个字符 'I':
字符'I'在字母表中的位置是: 25
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
加密计算: (25 + 17) % 26 = 16
加密后的字符是: U

处理第5个字符 'T':
字符'T'在字母表中的位置是: 4
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
加密计算: (4 + 38) % 26 = 16
加密后的字符是: U

处理第6个字符 'I':
字符'I'在字母表中的位置是: 25
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
加密计算: (25 + 45) % 26 = 18
加密后的字符是: P

处理第7个字符 'C':
字符'C'在字母表中的位置是: 7
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
加密计算: (7 + 45) % 26 = 0
加密后的字符是: J

最终加密结果: AVPUUPJ


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: Uppercase letters string without punctuation and spaces.
- Preparation:
    - 5x5 Gird(All row and column numbers are counted from 0.):
        - M Z S D P
        K N F L Q
        G A O X U
        W R Y V C
        B T E H I
        - Located at the first letter of all rows, MKGWB is the row header letter.
        - Located at the last letter of all columns, BTEHI is the bottom column letter.
- Encryption Steps:
    - Remove spaces and punctuation and the letter J from plain text and convert all letters to uppercase.
    - For each letter p in the plaintext.
        - Find the position of the letter p in the grid, and then find the corresponding row header and column footer characters accordingly.
        - Concatenate the row-header and column-bottom characters into a binary as the encrypted message for that letter p.
        - For example, if the letter p is H, it is in the 4th row, and the row header character is B; it is in the 3rd column, and the column bottom character is H, so the encrypted message is BH.

    Concatenate all the encrypted messages for the final ciphertext output.

clear text: pyritic encode into encrypted text: MIWEWTBIBTBIWI
clear text: nonculpability encode into encrypted text: KTGEKTWIGIKHMIGTBBBIKHBIBTWE
clear text: pyritic encode into encrypted text: MIWEWTBIBTBIWI
clear text: nervii encode into encrypted 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:
开始加密过程...
原始文本: nervii
预处理后的文本(移除空格标点,转大写,移除J): NERVII

逐字符加密:
字符 N 位于第2行第2列
-> 行首字符为K, 列尾字符为T
-> 加密为: KT
字符 E 位于第5行第3列
-> 行首字符为B, 列尾字符为E
-> 加密为: BE
字符 R 位于第4行第2列
-> 行首字符为W, 列尾字符为T
-> 加密为: WT
字符 V 位于第4行第4列
-> 行首字符为W, 列尾字符为H
-> 加密为: WH
字符 I 位于第5行第5列
-> 行首字符为B, 列尾字符为I
-> 加密为: BI
字符 I 位于第5行第5列
-> 行首字符为B, 列尾字符为I
-> 加密为: BI

最终加密结果: KTBEWTWHBIBI


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 without punctuation and spaces.
- Preparation:
    - 5x5 Gird(All row and column numbers are counted from 0.):
        - M Z S D P
        K N F L Q
        G A O X U
        W R Y V C
        B T E H I
        - Located at the first letter of all rows, MKGWB is the row header letter.
        - Located at the last letter of all columns, BTEHI is the bottom column letter.
- Encryption Steps:
    - Remove spaces and punctuation and the letter J from plain text and convert all letters to uppercase.
    - For each letter p in the plaintext.
        - Find the position of the letter p in the grid, and then find the corresponding row header and column footer characters accordingly.
        - Concatenate the row-header and column-bottom characters into a binary as the encrypted message for that letter p.
        - For example, if the letter p is H, it is in the 4th row, and the row header character is B; it is in the 3rd column, and the column bottom character is H, so the encrypted message is BH.

    Concatenate all the encrypted messages for the final ciphertext output.

plain text: nonculpability encrypt to 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: NONCULPABILITY

Output:
开始加密过程...
原始文本: nonculpability
预处理后的文本(移除空格标点,转大写,移除J): NONCULPABILITY

逐字符加密:
字符 N 位于第2行第2列
-> 行首字符为K, 列尾字符为T
-> 加密为: KT
字符 O 位于第3行第3列
-> 行首字符为G, 列尾字符为E
-> 加密为: GE
字符 N 位于第2行第2列
-> 行首字符为K, 列尾字符为T
-> 加密为: KT
字符 C 位于第4行第5列
-> 行首字符为W, 列尾字符为I
-> 加密为: WI
字符 U 位于第3行第5列
-> 行首字符为G, 列尾字符为I
-> 加密为: GI
字符 L 位于第2行第4列
-> 行首字符为K, 列尾字符为H
-> 加密为: KH
字符 P 位于第1行第5列
-> 行首字符为M, 列尾字符为I
-> 加密为: MI
字符 A 位于第3行第2列
-> 行首字符为G, 列尾字符为T
-> 加密为: GT
字符 B 位于第5行第1列
-> 行首字符为B, 列尾字符为B
-> 加密为: BB
字符 I 位于第5行第5列
-> 行首字符为B, 列尾字符为I
-> 加密为: BI
字符 L 位于第2行第4列
-> 行首字符为K, 列尾字符为H
-> 加密为: KH
字符 I 位于第5行第5列
-> 行首字符为B, 列尾字符为I
-> 加密为: BI
字符 T 位于第5行第2列
-> 行首字符为B, 列尾字符为T
-> 加密为: BT
字符 Y 位于第4行第3列
-> 行首字符为W, 列尾字符为E
-> 加密为: WE

最终加密结果: KTGEKTWIGIKHMIGTBBBIKHBIBTWE


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 without punctuation and spaces.
- Preparation:
    - 5x5 Gird(All row and column numbers are counted from 0.):
        - M Z S D P
        K N F L Q
        G A O X U
        W R Y V C
        B T E H I
        - Located at the first letter of all rows, MKGWB is the row header letter.
        - Located at the last letter of all columns, BTEHI is the bottom column letter.
- Encryption Steps:
    - Remove spaces and punctuation and the letter J from plain text and convert all letters to uppercase.
    - For each letter p in the plaintext.
        - Find the position of the letter p in the grid, and then find the corresponding row header and column footer characters accordingly.
        - Concatenate the row-header and column-bottom characters into a binary as the encrypted message for that letter p.
        - For example, if the letter p is H, it is in the 4th row, and the row header character is B; it is in the 3rd column, and the column bottom character is H, so the encrypted message is BH.

    Concatenate all the encrypted messages for the final ciphertext output.

clear text: nervii encode into cipher text: KTBEWTWHBIBI
clear text: nervii encode into cipher text: KTBEWTWHBIBI
clear text: pyritic encode 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...
原始文本: pyritic
预处理后的文本(移除空格标点,转大写,移除J): PYRITIC

逐字符加密:
字符 P 位于第1行第5列
-> 行首字符为M, 列尾字符为I
-> 加密为: MI
字符 Y 位于第4行第3列
-> 行首字符为W, 列尾字符为E
-> 加密为: WE
字符 R 位于第4行第2列
-> 行首字符为W, 列尾字符为T
-> 加密为: WT
字符 I 位于第5行第5列
-> 行首字符为B, 列尾字符为I
-> 加密为: BI
字符 T 位于第5行第2列
-> 行首字符为B, 列尾字符为T
-> 加密为: BT
字符 I 位于第5行第5列
-> 行首字符为B, 列尾字符为I
-> 加密为: BI
字符 C 位于第4行第5列
-> 行首字符为W, 列尾字符为I
-> 加密为: WI

最终加密结果: MIWEWTBIBTBIWI


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 without punctuation and spaces.
- Preparation:
    - 6x6 Matrix (The rows and columns in the matrix are counted from 0.)

        [['R', 'U', 'A', '0', 'Q', 'B'],
        ['D', '2', 'W', 'K', 'S', '1'],
        ['H', '4', '5', 'F', 'T', 'Z'],
        ['Y', 'C', 'G', 'X', '7', 'L'],
        ['9', '8', 'I', '3', 'P', 'N'],
        ['6', 'J', 'V', 'O', 'E', 'M']]

    - Encrypted Character Set
        - "ADFGVX"
- Encryption Steps:
    - For each character in the plaintext:
        - Find the number of rows and columns of that character in the 6X6 matrix. For example, A has 0 rows and 2 columns.
        - The position of each character in the encrypted character set is flagged as 0-6. Use the character at the corresponding position in the encrypted character set to replace the number of rows and columns to get the encrypted binary.
            - For example, the number of rows of A is 0, which corresponds to A in the encrypted character set, and the number of columns of A is 2, which corresponds to F in the encrypted character set. so the encrypted binary corresponding to A is (A,F).The encrypted binary is (A,F).
        - Read all rows of the encrypted binary, and then read all columns of the encrypted binary to get the final ciphertext.
            - For example, the encrypted binary is (A,F)(X,V), the last read is AXFV, so the final ciphertext is AXFV.

 **Decryption Rules:**

- Input:
    - Ciphertext: Uppercase letters  string without punctuation and spaces.
- Output:
    - Plaintext: Uppercase letters string without punctuation and spaces.
- Preparation:
    - 6x6 Matrix (Same as encryption)
    - Encrypted Character Set (Same as encryption)
- Decryption Steps:
    - Divide the ciphertext into two halves.
    - Take one letter from the upper half and one letter from the lower half each time as a decryption binary:
        - The position of each character in the encrypted character set is flagged as 0-6, and the position of the corresponding character in the encrypted character set is used to decrypt the number of rows and columns represented by the decrypted binary.
            - For example, the decrypted binary is the encrypted binary as (A,F), the position of A is 0 and the position of F is 2, so the number of rows is 0 and the number of columns is 2, obtaining (0, 2).
        - Use the number of rows and columns obtained to find the letter in the corresponding position in the 6x6 matrix as the character after decryption.
            - For example, the character at (0, 2) is A, and the character after AF decryption is A.
    - Concatenate all the decrypted characters to get the final plaintext.
encoded text: VGAVFVGVAAFVFD decrypt to plain text: pyritic
encoded text: VXAXVVXVAFFF decrypt to plain 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: VXAXVVXVAFFF

Output:
输入密文: VXAXVVXVAFFF

解密过程:
1. 将密文分成两半:
前半部分: VXAXVV
后半部分: XVAFFF

2. 将前后半部分配对并解密:
配对 (V,X) 对应矩阵位置 (4,5)，解密为字符: N
配对 (X,V) 对应矩阵位置 (5,4)，解密为字符: E
配对 (A,A) 对应矩阵位置 (0,0)，解密为字符: R
配对 (X,F) 对应矩阵位置 (5,2)，解密为字符: V
配对 (V,F) 对应矩阵位置 (4,2)，解密为字符: I
配对 (V,F) 对应矩阵位置 (4,2)，解密为字符: I

最终明文: NERVII


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 without punctuation and spaces.
- Preparation:
    - 6x6 Matrix (The rows and columns in the matrix are counted from 0.)

        [['R', 'U', 'A', '0', 'Q', 'B'],
        ['D', '2', 'W', 'K', 'S', '1'],
        ['H', '4', '5', 'F', 'T', 'Z'],
        ['Y', 'C', 'G', 'X', '7', 'L'],
        ['9', '8', 'I', '3', 'P', 'N'],
        ['6', 'J', 'V', 'O', 'E', 'M']]

    - Encrypted Character Set
        - "ADFGVX"
- Encryption Steps:
    - For each character in the plaintext:
        - Find the number of rows and columns of that character in the 6X6 matrix. For example, A has 0 rows and 2 columns.
        - The position of each character in the encrypted character set is flagged as 0-6. Use the character at the corresponding position in the encrypted character set to replace the number of rows and columns to get the encrypted binary.
            - For example, the number of rows of A is 0, which corresponds to A in the encrypted character set, and the number of columns of A is 2, which corresponds to F in the encrypted character set. so the encrypted binary corresponding to A is (A,F).The encrypted binary is (A,F).
        - Read all rows of the encrypted binary, and then read all columns of the encrypted binary to get the final ciphertext.
            - For example, the encrypted binary is (A,F)(X,V), the last read is AXFV, so the final ciphertext is AXFV.

 **Decryption Rules:**

- Input:
    - Ciphertext: Uppercase letters  string without punctuation and spaces.
- Output:
    - Plaintext: Uppercase letters string without punctuation and spaces.
- Preparation:
    - 6x6 Matrix (Same as encryption)
    - Encrypted Character Set (Same as encryption)
- Decryption Steps:
    - Divide the ciphertext into two halves.
    - Take one letter from the upper half and one letter from the lower half each time as a decryption binary:
        - The position of each character in the encrypted character set is flagged as 0-6, and the position of the corresponding character in the encrypted character set is used to decrypt the number of rows and columns represented by the decrypted binary.
            - For example, the decrypted binary is the encrypted binary as (A,F), the position of A is 0 and the position of F is 2, so the number of rows is 0 and the number of columns is 2, obtaining (0, 2).
        - Use the number of rows and columns obtained to find the letter in the corresponding position in the 6x6 matrix as the character after decryption.
            - For example, the character at (0, 2) is A, and the character after AF decryption is A.
    - Concatenate all the decrypted characters to get the final plaintext.
encrypted text: VXVGAGVAAVGVFGXGXDDXVFXFXFVA decode into clear 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: VXVGAGVAAVGVFGXGXDDXVFXFXFVA

Output:
输入密文: VXVGAGVAAVGVFGXGXDDXVFXFXFVA

解密过程:
1. 将密文分成两半:
前半部分: VXVGAGVAAVGVFG
后半部分: XGXDDXVFXFXFVA

2. 将前后半部分配对并解密:
配对 (V,X) 对应矩阵位置 (4,5)，解密为字符: N
配对 (X,G) 对应矩阵位置 (5,3)，解密为字符: O
配对 (V,X) 对应矩阵位置 (4,5)，解密为字符: N
配对 (G,D) 对应矩阵位置 (3,1)，解密为字符: C
配对 (A,D) 对应矩阵位置 (0,1)，解密为字符: U
配对 (G,X) 对应矩阵位置 (3,5)，解密为字符: L
配对 (V,V) 对应矩阵位置 (4,4)，解密为字符: P
配对 (A,F) 对应矩阵位置 (0,2)，解密为字符: A
配对 (A,X) 对应矩阵位置 (0,5)，解密为字符: B
配对 (V,F) 对应矩阵位置 (4,2)，解密为字符: I
配对 (G,X) 对应矩阵位置 (3,5)，解密为字符: L
配对 (V,F) 对应矩阵位置 (4,2)，解密为字符: I
配对 (F,V) 对应矩阵位置 (2,4)，解密为字符: T
配对 (G,A) 对应矩阵位置 (3,0)，解密为字符: Y

最终明文: NONCULPABILITY


Prompt:
You are an excellent cipher decoder, please refer to the following examples and information to decode the ciphertext.
**Encryption Rules:**

- Input:
    - Plaintext: Uppercase letters string without punctuation and spaces.
- Output:
    - Ciphertext:  Uppercase letters  string without punctuation and spaces.
- Preparation:
    - 6x6 Matrix (The rows and columns in the matrix are counted from 0.)

        [['R', 'U', 'A', '0', 'Q', 'B'],
        ['D', '2', 'W', 'K', 'S', '1'],
        ['H', '4', '5', 'F', 'T', 'Z'],
        ['Y', 'C', 'G', 'X', '7', 'L'],
        ['9', '8', 'I', '3', 'P', 'N'],
        ['6', 'J', 'V', 'O', 'E', 'M']]

    - Encrypted Character Set
        - "ADFGVX"
- Encryption Steps:
    - For each character in the plaintext:
        - Find the number of rows and columns of that character in the 6X6 matrix. For example, A has 0 rows and 2 columns.
        - The position of each character in the encrypted character set is flagged as 0-6. Use the character at the corresponding position in the encrypted character set to replace the number of rows and columns to get the encrypted binary.
            - For example, the number of rows of A is 0, which corresponds to A in the encrypted character set, and the number of columns of A is 2, which corresponds to F in the encrypted character set. so the encrypted binary corresponding to A is (A,F).The encrypted binary is (A,F).
        - Read all rows of the encrypted binary, and then read all columns of the encrypted binary to get the final ciphertext.
            - For example, the encrypted binary is (A,F)(X,V), the last read is AXFV, so the final ciphertext is AXFV.

 **Decryption Rules:**

- Input:
    - Ciphertext: Uppercase letters  string without punctuation and spaces.
- Output:
    - Plaintext: Uppercase letters string without punctuation and spaces.
- Preparation:
    - 6x6 Matrix (Same as encryption)
    - Encrypted Character Set (Same as encryption)
- Decryption Steps:
    - Divide the ciphertext into two halves.
    - Take one letter from the upper half and one letter from the lower half each time as a decryption binary:
        - The position of each character in the encrypted character set is flagged as 0-6, and the position of the corresponding character in the encrypted character set is used to decrypt the number of rows and columns represented by the decrypted binary.
            - For example, the decrypted binary is the encrypted binary as (A,F), the position of A is 0 and the position of F is 2, so the number of rows is 0 and the number of columns is 2, obtaining (0, 2).
        - Use the number of rows and columns obtained to find the letter in the corresponding position in the 6x6 matrix as the character after decryption.
            - For example, the character at (0, 2) is A, and the character after AF decryption is A.
    - Concatenate all the decrypted characters to get the final plaintext.
cipher text: VGAVFVGVAAFVFD decrypt into clear 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: VGAVFVGVAAFVFD

Output:
输入密文: VGAVFVGVAAFVFD

解密过程:
1. 将密文分成两半:
前半部分: VGAVFVG
后半部分: VAAFVFD

2. 将前后半部分配对并解密:
配对 (V,V) 对应矩阵位置 (4,4)，解密为字符: P
配对 (G,A) 对应矩阵位置 (3,0)，解密为字符: Y
配对 (A,A) 对应矩阵位置 (0,0)，解密为字符: R
配对 (V,F) 对应矩阵位置 (4,2)，解密为字符: I
配对 (F,V) 对应矩阵位置 (2,4)，解密为字符: T
配对 (V,F) 对应矩阵位置 (4,2)，解密为字符: I
配对 (G,D) 对应矩阵位置 (3,1)，解密为字符: C

最终明文: 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:
    - 暗文: 大写字母字符串
- 准备:
以下是经过整理后的文字：

---

### 5x5 网格布局

共有四个5x5的字符网格，每个网格代表了不同的排列方式。

1. **网格1 原始网格**

这是最初的字符排列，按照特定顺序组织如下：

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

2. **网格2 ECHO 网格**

该网格根据"ECHO"这个词进行了重新排列：

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

3. **网格3 VORTEX 网格**

此网格基于"VORTEX"一词进行了独特的字符重组：

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

4. **网格4 重复原始网格**

最后一个网格与第一个原始网格完全相同，没有进行任何改变：

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

每个网格展示了不同主题词下字符的独特排列。
- 加密步骤:
    - 清理原文，移除空格和非字母字符，移除字母Q，转换为大写
    - 如果原文长度为奇数，添加字母'X'使其成为偶数
    - 将处理后的原文分成两个字母一组
    - 对于每组两个字母p1,p2:
        - 在网格1和网格4中找到第一个字母和第二个字母的位置
        - 在网格2和网格3中找到这两个位置对应的字母，用这两个字母作为该组的加密结果
    - 连接所有加密后的字母组形成最终暗文
解密算法:
解密规则:
- 输入:
    - 暗文: 大写字母字符串
- 输出:
    - 原文: 大写字母字符串
- 准备:
    - 四个5x5网格(与加密相同)
- 解密步骤(与加密步骤相反):
    - 清理暗文，移除空格和非字母字符，转换为大写
    - 将处理后的暗文分成两个字母一组
    - 对于每组两个字母c1,c2:
        - 在网格2和网格3中找到第一个字母和第二个字母的位置
        - 在网格1和网格4中找到这两个位置对应的字母，用这两个字母作为该组的解密结果
    - 连接所有解密后的字母组形成最终原文
暗文: BIDYGYSH 解密成原文: pyritic
暗文: OPDFYY 解密成原文: ? 请一步一步完成，确保过程详细严谨，结果正确。

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

Input: OPDFYY

Output:
开始解密过程...
加密文本: OPDFYY

开始逐对字母解密:

处理加密字母对: OP
第一个字母 O 在ECHO表格中的位置: (0, 3)
第二个字母 P 在VORTEX表格中的位置: (3, 4)
解密后的字母对: NE

处理加密字母对: DF
第一个字母 D 在ECHO表格中的位置: (1, 1)
第二个字母 F 在VORTEX表格中的位置: (2, 0)
解密后的字母对: RV

处理加密字母对: YY
第一个字母 Y 在ECHO表格中的位置: (4, 3)
第二个字母 Y 在VORTEX表格中的位置: (4, 3)
解密后的字母对: II

最终解密结果: NERVII


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:
    - 密文: 大写字母字符串
- 准备:
以下是经过整理后的文字：

---

### 5x5 网格布局

共有四个5x5的字符网格，每个网格代表了不同的排列方式。

1. **网格1 原始网格**

这是最初的字符排列，按照特定顺序组织如下：

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

2. **网格2 ECHO 网格**

该网格根据"ECHO"这个词进行了重新排列：

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

3. **网格3 VORTEX 网格**

此网格基于"VORTEX"一词进行了独特的字符重组：

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

4. **网格4 重复原始网格**

最后一个网格与第一个原始网格完全相同，没有进行任何改变：

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

每个网格展示了不同主题词下字符的独特排列。
- 加密步骤:
    - 清理原始信息，移除空格和非字母字符，移除字母Q，转换为大写
    - 如果原始信息长度为奇数，添加字母'X'使其成为偶数
    - 将处理后的原始信息分成两个字母一组
    - 对于每组两个字母p1,p2:
        - 在网格1和网格4中找到第一个字母和第二个字母的位置
        - 在网格2和网格3中找到这两个位置对应的字母，用这两个字母作为该组的加密结果
    - 连接所有加密后的字母组形成最终密文
解密算法:
解密规则:
- 输入:
    - 密文: 大写字母字符串
- 输出:
    - 原始信息: 大写字母字符串
- 准备:
    - 四个5x5网格(与加密相同)
- 解密步骤(与加密步骤相反):
    - 清理密文，移除空格和非字母字符，转换为大写
    - 将处理后的密文分成两个字母一组
    - 对于每组两个字母c1,c2:
        - 在网格2和网格3中找到第一个字母和第二个字母的位置
        - 在网格1和网格4中找到这两个位置对应的字母，用这两个字母作为该组的解密结果
    - 连接所有解密后的字母组形成最终原始信息
密文: BIDYGYSH 解密成原始信息: pyritic
密文: BIDYGYSH 解密成原始信息: pyritic
密文: BIDYGYSH 解密成原始信息: pyritic
密文: OEOMIOBKRYCYGI 解密成原始信息: ? 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: OEOMIOBKRYCYGI

Output:
开始解密过程...
加密文本: OEOMIOBKRYCYGI

开始逐对字母解密:

处理加密字母对: OE
第一个字母 O 在ECHO表格中的位置: (0, 3)
第二个字母 E 在VORTEX表格中的位置: (0, 4)
解密后的字母对: NO

处理加密字母对: OM
第一个字母 O 在ECHO表格中的位置: (0, 3)
第二个字母 M 在VORTEX表格中的位置: (3, 2)
解密后的字母对: NC

处理加密字母对: IO
第一个字母 I 在ECHO表格中的位置: (1, 4)
第二个字母 O 在VORTEX表格中的位置: (0, 1)
解密后的字母对: UL

处理加密字母对: BK
第一个字母 B 在ECHO表格中的位置: (1, 0)
第二个字母 K 在VORTEX表格中的位置: (3, 0)
解密后的字母对: PA

处理加密字母对: RY
第一个字母 R 在ECHO表格中的位置: (3, 1)
第二个字母 Y 在VORTEX表格中的位置: (4, 3)
解密后的字母对: BI

处理加密字母对: CY
第一个字母 C 在ECHO表格中的位置: (0, 1)
第二个字母 Y 在VORTEX表格中的位置: (4, 3)
解密后的字母对: LI

处理加密字母对: GI
第一个字母 G 在ECHO表格中的位置: (1, 3)
第二个字母 I 在VORTEX表格中的位置: (2, 3)
解密后的字母对: TY

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:
    - 原始信息: 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:
以下是经过整理后的文字：

---

### 5x5 网格布局

共有四个5x5的字符网格，每个网格代表了不同的排列方式。

1. **网格1 原始网格**

这是最初的字符排列，按照特定顺序组织如下：

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

2. **网格2 ECHO 网格**

该网格根据"ECHO"这个词进行了重新排列：

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

3. **网格3 VORTEX 网格**

此网格基于"VORTEX"一词进行了独特的字符重组：

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

4. **网格4 重复原始网格**

最后一个网格与第一个原始网格完全相同，没有进行任何改变：

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

每个网格展示了不同主题词下字符的独特排列。
- 加密步骤:
    - 清理原始信息，移除空格和非字母字符，移除字母Q，转换为大写
    - 如果原始信息长度为奇数，添加字母'X'使其成为偶数
    - 将处理后的原始信息分成两个字母一组
    - 对于每组两个字母p1,p2:
        - 在网格1和网格4中找到第一个字母和第二个字母的位置
        - 在网格2和网格3中找到这两个位置对应的字母，用这两个字母作为该组的加密结果
    - 连接所有加密后的字母组形成最终加密文本
解密算法:
解密规则:
- 输入:
    - 加密文本: 大写字母字符串
- 输出:
    - 原始信息: 大写字母字符串
- 准备:
    - 四个5x5网格(与加密相同)
- 解密步骤(与加密步骤相反):
    - 清理加密文本，移除空格和非字母字符，转换为大写
    - 将处理后的加密文本分成两个字母一组
    - 对于每组两个字母c1,c2:
        - 在网格2和网格3中找到第一个字母和第二个字母的位置
        - 在网格1和网格4中找到这两个位置对应的字母，用这两个字母作为该组的解密结果
    - 连接所有解密后的字母组形成最终原始信息
加密文本: BIDYGYSH 解密为原始信息是什么？

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

Input: BIDYGYSH

Output:
开始解密过程...
加密文本: BIDYGYSH

开始逐对字母解密:

处理加密字母对: BI
第一个字母 B 在ECHO表格中的位置: (1, 0)
第二个字母 I 在VORTEX表格中的位置: (2, 3)
解密后的字母对: PY

处理加密字母对: DY
第一个字母 D 在ECHO表格中的位置: (1, 1)
第二个字母 Y 在VORTEX表格中的位置: (4, 3)
解密后的字母对: RI

处理加密字母对: GY
第一个字母 G 在ECHO表格中的位置: (1, 3)
第二个字母 Y 在VORTEX表格中的位置: (4, 3)
解密后的字母对: TI

处理加密字母对: SH
第一个字母 S 在ECHO表格中的位置: (3, 2)
第二个字母 H 在VORTEX表格中的位置: (2, 2)
解密后的字母对: CX

最终解密结果: PYRITICX


Prompt:
您是一位杰出的密文加密专家，请参考以下案例和信息进行加密操作。
非加密信息: pyritic 密钥或额外参数: {'shift': 1} 编码为加密文本: qzsjujd
非加密信息: pyritic 密钥或额外参数: {'shift': 1} 编码为加密文本: qzsjujd
非加密信息: nonculpability 密钥或额外参数: {'shift': 15} 编码为加密文本: cdcrjaepqxaxin
非加密信息: nonculpability 密钥或额外参数: {'shift': 15} 编码为加密文本: cdcrjaepqxaxin
非加密信息: nervii 密钥或额外参数: {'shift': 7} 编码为加密文本: ? 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: 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:
您的任务是使用相应的算法将敏感信息转换为不可读的形式，以保障其传输过程中的安全性。
明文: pyritic 密钥或额外参数: {'shift': 1} 加密成为隐文: qzsjujd
明文: pyritic 密钥或额外参数: {'shift': 1} 加密成为隐文: qzsjujd
明文: pyritic 密钥或额外参数: {'shift': 1} 加密成为隐文: qzsjujd
明文: nervii 密钥或额外参数: {'shift': 7} 加密成为隐文: ulycpp
明文: pyritic 密钥或额外参数: {'shift': 1} 加密成为隐文: qzsjujd
明文: nonculpability 密钥或额外参数: {'shift': 15} 加密成为隐文是什么？ 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: 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:
您的任务是使用相应的算法将敏感信息转换为不可读的形式，以保障其传输过程中的安全性。
原文: nonculpability 密钥或额外参数: {'shift': 15} 加密为隐文: cdcrjaepqxaxin
原文: nonculpability 密钥或额外参数: {'shift': 15} 加密为隐文: cdcrjaepqxaxin
原文: nervii 密钥或额外参数: {'shift': 7} 加密为隐文: ulycpp
原文: nonculpability 密钥或额外参数: {'shift': 15} 加密为隐文: cdcrjaepqxaxin
原文: nervii 密钥或额外参数: {'shift': 7} 加密为隐文: ulycpp
原文: pyritic 密钥或额外参数: {'shift': 1} 加密为隐文: ? 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: 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:
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:
解题方案：
1. 构建3个矩阵：默认字母表矩阵M_T，加密矩阵M1和M2（由str1和str2生成）。
2. 将非加密信息文本转换为大写字母，仅保留字母字符并分成两个字母一组。
3. 对每对字母进行加密：
   - 查找每个字母在默认矩阵中的位置，交换其y坐标，
   - 根据新坐标从M1和M2矩阵中取出加密后的字母对。
4. 输出加密后的文本。
解密算法:
解题方案：
1. 构建3个矩阵：默认字母表矩阵M_T，加密矩阵M1和M2（由str1和str2生成）。
2. 将密文按两个字母一组分组。
3. 对每对加密字母进行解密：
   - 在M1和M2中找到加密字母的位置，交换列坐标
   - 从默认矩阵中获取对应的非加密信息字母。
4. 输出解密后的文本。
密文: JK FI LL KS CE GE NY 密钥或额外参数: {'str1': 'trichophytia', 'str2': 'sphinges'} 解密成为非加密信息: nonculpability
密文: JK FI LL KS CE GE NY 密钥或额外参数: {'str1': 'trichophytia', 'str2': 'sphinges'} 解密成为非加密信息: nonculpability
密文: RV RB RD WX 密钥或额外参数: {'str1': 'sawbill', 'str2': 'houri'} 解密成为非加密信息: pyritic
密文: JV KT CA 密钥或额外参数: {'str1': 'parametral', 'str2': 'quiverful'} 解密成为非加密信息: ? 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: JV KT CA

Output:
解题方案：
1. 构建3个矩阵：默认字母表矩阵M_T，加密矩阵M1和M2（由str1和str2生成）。
2. 将密文按两个字母一组分组。
3. 对每对加密字母进行解密：
   - 在M1和M2中找到加密字母的位置，交换列坐标
   - 从默认矩阵中获取对应的明文字母。
4. 输出解密后的文本。

Step 1: 创建默认字母表矩阵 M_T (去除Q)：
A B C D E
F G H I J
K L M N O
P R S T U
V W X Y Z

Step 2: 使用密钥构建加密矩阵 M1 和 M2
M1 矩阵 (根据密钥 str1):
P A R M E
T L B C D
F G H I J
K N O S U
V W X Y Z

M2 矩阵 (根据密钥 str2):
Q U I V E
R F L A B
C D G H J
K M N O P
S T W X Y

Step 3: 将密文分成两个字母一组: ['JV', 'KT', 'CA']

Step 4.1: 解密字母对 JV
  - J 在 M1 中的位置: (2, 4)
  - V 在 M2 中的位置: (0, 3)
  - 交换列索引后新位置: (2, 3) 和 (0, 4)
  - 在 M_T 中查找位置 (2, 3) 的字母: N
  - 在 M_T 中查找位置 (0, 4) 的字母: E
  - 解密结果为: NE

Step 4.2: 解密字母对 KT
  - K 在 M1 中的位置: (3, 0)
  - T 在 M2 中的位置: (4, 1)
  - 交换列索引后新位置: (3, 1) 和 (4, 0)
  - 在 M_T 中查找位置 (3, 1) 的字母: R
  - 在 M_T 中查找位置 (4, 0) 的字母: V
  - 解密结果为: RV

Step 4.3: 解密字母对 CA
  - C 在 M1 中的位置: (1, 3)
  - A 在 M2 中的位置: (1, 3)
  - 交换列索引后新位置: (1, 3) 和 (1, 3)
  - 在 M_T 中查找位置 (1, 3) 的字母: I
  - 在 M_T 中查找位置 (1, 3) 的字母: I
  - 解密结果为: II

解密完成！最终解密结果为: NERVII


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:
解题方案：
1. 构建3个矩阵：默认字母表矩阵M_T，加密矩阵M1和M2（由str1和str2生成）。
2. 将原始信息文本转换为大写字母，仅保留字母字符并分成两个字母一组。
3. 对每对字母进行加密：
   - 查找每个字母在默认矩阵中的位置，交换其y坐标，
   - 根据新坐标从M1和M2矩阵中取出加密后的字母对。
4. 输出加密后的文本。
解密算法:
解题方案：
1. 构建3个矩阵：默认字母表矩阵M_T，加密矩阵M1和M2（由str1和str2生成）。
2. 将加密文本按两个字母一组分组。
3. 对每对加密字母进行解密：
   - 在M1和M2中找到加密字母的位置，交换列坐标
   - 从默认矩阵中获取对应的原始信息字母。
4. 输出解密后的文本。
加密文本: JV KT CA 密钥或额外参数: {'str1': 'parametral', 'str2': 'quiverful'} 解密为原始信息: nervii
加密文本: JV KT CA 密钥或额外参数: {'str1': 'parametral', 'str2': 'quiverful'} 解密为原始信息: nervii
加密文本: JV KT CA 密钥或额外参数: {'str1': 'parametral', 'str2': 'quiverful'} 解密为原始信息: nervii
加密文本: JK FI LL KS CE GE NY 密钥或额外参数: {'str1': 'trichophytia', 'str2': 'sphinges'} 解密为原始信息是什么？ 请一步一步完成，确保过程详细严谨，结果正确。

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

Input: JK FI LL KS CE GE NY

Output:
解题方案：
1. 构建3个矩阵：默认字母表矩阵M_T，加密矩阵M1和M2（由str1和str2生成）。
2. 将密文按两个字母一组分组。
3. 对每对加密字母进行解密：
   - 在M1和M2中找到加密字母的位置，交换列坐标
   - 从默认矩阵中获取对应的明文字母。
4. 输出解密后的文本。

Step 1: 创建默认字母表矩阵 M_T (去除Q)：
A B C D E
F G H I J
K L M N O
P R S T U
V W X Y Z

Step 2: 使用密钥构建加密矩阵 M1 和 M2
M1 矩阵 (根据密钥 str1):
T R I C H
O P Y A B
D E F G J
K L M N S
U V W X Z

M2 矩阵 (根据密钥 str2):
S P H I N
G E A B C
D F J K L
M O R T U
V W X Y Z

Step 3: 将密文分成两个字母一组: ['JK', 'FI', 'LL', 'KS', 'CE', 'GE', 'NY']

Step 4.1: 解密字母对 JK
  - J 在 M1 中的位置: (2, 4)
  - K 在 M2 中的位置: (2, 3)
  - 交换列索引后新位置: (2, 3) 和 (2, 4)
  - 在 M_T 中查找位置 (2, 3) 的字母: N
  - 在 M_T 中查找位置 (2, 4) 的字母: O
  - 解密结果为: NO

Step 4.2: 解密字母对 FI
  - F 在 M1 中的位置: (2, 2)
  - I 在 M2 中的位置: (0, 3)
  - 交换列索引后新位置: (2, 3) 和 (0, 2)
  - 在 M_T 中查找位置 (2, 3) 的字母: N
  - 在 M_T 中查找位置 (0, 2) 的字母: C
  - 解密结果为: NC

Step 4.3: 解密字母对 LL
  - L 在 M1 中的位置: (3, 1)
  - L 在 M2 中的位置: (2, 4)
  - 交换列索引后新位置: (3, 4) 和 (2, 1)
  - 在 M_T 中查找位置 (3, 4) 的字母: U
  - 在 M_T 中查找位置 (2, 1) 的字母: L
  - 解密结果为: UL

Step 4.4: 解密字母对 KS
  - K 在 M1 中的位置: (3, 0)
  - S 在 M2 中的位置: (0, 0)
  - 交换列索引后新位置: (3, 0) 和 (0, 0)
  - 在 M_T 中查找位置 (3, 0) 的字母: P
  - 在 M_T 中查找位置 (0, 0) 的字母: A
  - 解密结果为: PA

Step 4.5: 解密字母对 CE
  - C 在 M1 中的位置: (0, 3)
  - E 在 M2 中的位置: (1, 1)
  - 交换列索引后新位置: (0, 1) 和 (1, 3)
  - 在 M_T 中查找位置 (0, 1) 的字母: B
  - 在 M_T 中查找位置 (1, 3) 的字母: I
  - 解密结果为: BI

Step 4.6: 解密字母对 GE
  - G 在 M1 中的位置: (2, 3)
  - E 在 M2 中的位置: (1, 1)
  - 交换列索引后新位置: (2, 1) 和 (1, 3)
  - 在 M_T 中查找位置 (2, 1) 的字母: L
  - 在 M_T 中查找位置 (1, 3) 的字母: I
  - 解密结果为: LI

Step 4.7: 解密字母对 NY
  - N 在 M1 中的位置: (3, 3)
  - Y 在 M2 中的位置: (4, 3)
  - 交换列索引后新位置: (3, 3) 和 (4, 3)
  - 在 M_T 中查找位置 (3, 3) 的字母: T
  - 在 M_T 中查找位置 (4, 3) 的字母: Y
  - 解密结果为: TY

解密完成！最终解密结果为: 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:
解题方案：
1. 构建3个矩阵：默认字母表矩阵M_T，加密矩阵M1和M2（由str1和str2生成）。
2. 将初始文本文本转换为大写字母，仅保留字母字符并分成两个字母一组。
3. 对每对字母进行加密：
   - 查找每个字母在默认矩阵中的位置，交换其y坐标，
   - 根据新坐标从M1和M2矩阵中取出加密后的字母对。
4. 输出加密后的文本。
解密算法:
解题方案：
1. 构建3个矩阵：默认字母表矩阵M_T，加密矩阵M1和M2（由str1和str2生成）。
2. 将加密文本按两个字母一组分组。
3. 对每对加密字母进行解密：
   - 在M1和M2中找到加密字母的位置，交换列坐标
   - 从默认矩阵中获取对应的初始文本字母。
4. 输出解密后的文本。
加密文本: JK FI LL KS CE GE NY 密钥或额外参数: {'str1': 'trichophytia', 'str2': 'sphinges'} 解密成为初始文本: nonculpability
加密文本: JK FI LL KS CE GE NY 密钥或额外参数: {'str1': 'trichophytia', 'str2': 'sphinges'} 解密成为初始文本: nonculpability
加密文本: JK FI LL KS CE GE NY 密钥或额外参数: {'str1': 'trichophytia', 'str2': 'sphinges'} 解密成为初始文本: nonculpability
加密文本: RV RB RD WX 密钥或额外参数: {'str1': 'sawbill', 'str2': 'houri'} 解密成为初始文本是什么？ 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: RV RB RD WX

Output:
解题方案：
1. 构建3个矩阵：默认字母表矩阵M_T，加密矩阵M1和M2（由str1和str2生成）。
2. 将密文按两个字母一组分组。
3. 对每对加密字母进行解密：
   - 在M1和M2中找到加密字母的位置，交换列坐标
   - 从默认矩阵中获取对应的明文字母。
4. 输出解密后的文本。

Step 1: 创建默认字母表矩阵 M_T (去除Q)：
A B C D E
F G H I J
K L M N O
P R S T U
V W X Y Z

Step 2: 使用密钥构建加密矩阵 M1 和 M2
M1 矩阵 (根据密钥 str1):
S A W B I
L C D E F
G H J K M
N O P R T
U V X Y Z

M2 矩阵 (根据密钥 str2):
H O U R I
A B C D E
F G J K L
M N P S T
V W X Y Z

Step 3: 将密文分成两个字母一组: ['RV', 'RB', 'RD', 'WX']

Step 4.1: 解密字母对 RV
  - R 在 M1 中的位置: (3, 3)
  - V 在 M2 中的位置: (4, 0)
  - 交换列索引后新位置: (3, 0) 和 (4, 3)
  - 在 M_T 中查找位置 (3, 0) 的字母: P
  - 在 M_T 中查找位置 (4, 3) 的字母: Y
  - 解密结果为: PY

Step 4.2: 解密字母对 RB
  - R 在 M1 中的位置: (3, 3)
  - B 在 M2 中的位置: (1, 1)
  - 交换列索引后新位置: (3, 1) 和 (1, 3)
  - 在 M_T 中查找位置 (3, 1) 的字母: R
  - 在 M_T 中查找位置 (1, 3) 的字母: I
  - 解密结果为: RI

Step 4.3: 解密字母对 RD
  - R 在 M1 中的位置: (3, 3)
  - D 在 M2 中的位置: (1, 3)
  - 交换列索引后新位置: (3, 3) 和 (1, 3)
  - 在 M_T 中查找位置 (3, 3) 的字母: T
  - 在 M_T 中查找位置 (1, 3) 的字母: I
  - 解密结果为: TI

Step 4.4: 解密字母对 WX
  - W 在 M1 中的位置: (0, 2)
  - X 在 M2 中的位置: (4, 2)
  - 交换列索引后新位置: (0, 2) 和 (4, 2)
  - 在 M_T 中查找位置 (0, 2) 的字母: C
  - 在 M_T 中查找位置 (4, 2) 的字母: X
  - 解密结果为: CX

解密完成！最终解密结果为: PYRITICX


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.
cipher text: OCAX..C decode to clear text: pyritic
cipher text: NFJBXZUHXEJCCY decode to clear text: nonculpability
cipher text: NFJBXZUHXEJCCY decode to clear text: nonculpability
cipher text: MSQIQR decode to clear 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: MSQIQR

Output:
步骤1: 将密文转换为大写: MSQIQR
步骤2: 将每个字母转换为坐标: 212311223133223233
步骤3: 计算列数: 6
步骤4: 将坐标组织成3xN的矩阵:
212311
223133
223233
步骤5: 转置矩阵:
222
122
233
312
133
133
立方体布局:
第1层:
A B C
D E F
G H I

第2层:
J K L
M N O
P Q R

第3层:
S T U
V W X
Y Z .

步骤6: 组合转置后的坐标: [(2, 2, 2), (1, 2, 2), (2, 3, 3), (3, 1, 2), (1, 3, 3), (1, 3, 3)]
步骤7: 将坐标转换回字母: NERVII


Prompt:
Please decode the ciphertext according to the examples and the given information.
encoded text: MSQIQR decode to clear text: nervii
encoded text: MSQIQR decode to clear text: nervii
encoded text: MSQIQR decode to clear text: nervii
encoded text: OCAX..C decode to clear text: pyritic
encoded text: NFJBXZUHXEJCCY decode to clear 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: NFJBXZUHXEJCCY

Output:
步骤1: 将密文转换为大写: NFJBXZUHXEJCCY
步骤2: 将每个字母转换为坐标: 222132211121332323331123332122211131131313
步骤3: 计算列数: 14
步骤4: 将坐标组织成3xN的矩阵:
22213221112133
23233311233321
22211131131313
步骤5: 转置矩阵:
222
232
222
131
331
231
213
111
121
133
231
133
321
313
立方体布局:
第1层:
A B C
D E F
G H I

第2层:
J K L
M N O
P Q R

第3层:
S T U
V W X
Y Z .

步骤6: 组合转置后的坐标: [(2, 2, 2), (2, 3, 2), (2, 2, 2), (1, 3, 1), (3, 3, 1), (2, 3, 1), (2, 1, 3), (1, 1, 1), (1, 2, 1), (1, 3, 3), (2, 3, 1), (1, 3, 3), (3, 2, 1), (3, 1, 3)]
步骤7: 将坐标转换回字母: NONCULPABILITY


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.
encoded text: MSQIQR decrypt to original information: nervii
encoded text: MSQIQR decrypt to original information: nervii
encoded text: MSQIQR decrypt to original information: nervii
encoded text: MSQIQR decrypt to original information: nervii
encoded text: MSQIQR decrypt to original information: nervii
encoded text: OCAX..C decrypt to original information: ? 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: OCAX..C

Output:
步骤1: 将密文转换为大写: OCAX..C
步骤2: 将每个字母转换为坐标: 232131111332333333131
步骤3: 计算列数: 7
步骤4: 将坐标组织成3xN的矩阵:
2321311
1133233
3333131
步骤5: 转置矩阵:
213
313
233
133
321
133
131
立方体布局:
第1层:
A B C
D E F
G H I

第2层:
J K L
M N O
P Q R

第3层:
S T U
V W X
Y Z .

步骤6: 组合转置后的坐标: [(2, 1, 3), (3, 1, 3), (2, 3, 3), (1, 3, 3), (3, 2, 1), (1, 3, 3), (1, 3, 1)]
步骤7: 将坐标转换回字母: PYRITIC


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:  Hexadecimal string (and letters a-e in lower case).
- Preparation:
    - Key(SHA hash value)
        - Performing SHA-256 on "SECRET_KEY" results in '73ef2a4edd7a7fbf07fd5f6faf99674dc0c25a025fd74c221f4c35849e5c0fb3'.
- Encryption Steps:
    - Converts a plaintext string into a sequence of bytes (ASCII encoding).
    - Each byte is encrypted with the corresponding byte of the key using a bitwise different-or (XOR) operation. Repeatedly recycle the key so that the key length is the same as the number of plaintext bytes.
    - The encrypted byte sequence is converted to a hexadecimal string for ciphertext output.

 **Decryption Rules:**

- Input:
    - Ciphertext: Hexadecimal string (and letters a-e in lower case).
- Output:
    - Plaintext: Uppercase letters string without punctuation and spaces.
- Preparation:
    - Key (Same as encryption,is a hashed hash obtained by SHA_256)
- Decryption Steps:
    - Converts the ciphertext string into a sequence of bytes.
    - Each byte is decrypted (the decryption process is the same as the encryption process) with the corresponding byte of the key using a bitwise different-or (XOR) operation.
    - The decrypted byte sequence is converted to a plaintext string.
encrypted text: 3da0640d88362ffe45b41326fbc0 decode to plain text: nonculpability
encrypted text: 23b6780789333c decode to plain text: pyritic
encrypted text: 3daa78189433 decode to plain text is: ?  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...
加密的十六进制文本: 3daa78189433
生成的SHA-256密钥: 73ef2a4edd7a7fbf07fd5f6faf99674dc0c25a025fd74c221f4c35849e5c0fb3
十六进制转换为字节序列: b'=\xaax\x18\x943'
开始XOR解密...
解密后的字节序列: b'NERVII'
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:  Hexadecimal string (and letters a-e in lower case).
- Preparation:
    - Key(SHA hash value)
        - Performing SHA-256 on "SECRET_KEY" results in '73ef2a4edd7a7fbf07fd5f6faf99674dc0c25a025fd74c221f4c35849e5c0fb3'.
- Encryption Steps:
    - Converts a plaintext string into a sequence of bytes (ASCII encoding).
    - Each byte is encrypted with the corresponding byte of the key using a bitwise different-or (XOR) operation. Repeatedly recycle the key so that the key length is the same as the number of plaintext bytes.
    - The encrypted byte sequence is converted to a hexadecimal string for ciphertext output.

 **Decryption Rules:**

- Input:
    - Ciphertext: Hexadecimal string (and letters a-e in lower case).
- Output:
    - Plaintext: Uppercase letters string without punctuation and spaces.
- Preparation:
    - Key (Same as encryption,is a hashed hash obtained by SHA_256)
- Decryption Steps:
    - Converts the ciphertext string into a sequence of bytes.
    - Each byte is decrypted (the decryption process is the same as the encryption process) with the corresponding byte of the key using a bitwise different-or (XOR) operation.
    - The decrypted byte sequence is converted to a plaintext string.
cipher text: 23b6780789333c decrypt into clear text: pyritic
cipher text: 3da0640d88362ffe45b41326fbc0 decrypt into clear 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
```

Output:
开始解密过程...
加密的十六进制文本: 3da0640d88362ffe45b41326fbc0
生成的SHA-256密钥: 73ef2a4edd7a7fbf07fd5f6faf99674dc0c25a025fd74c221f4c35849e5c0fb3
十六进制转换为字节序列: b'=\xa0d\r\x886/\xfeE\xb4\x13&\xfb\xc0'
开始XOR解密...
解密后的字节序列: b'NONCULPABILITY'
最终解密结果: NONCULPABILITY


Prompt:
You are an excellent cipher decoder, please refer to the following examples and information to decode the ciphertext.
**Encryption Rules:**

- Input:
    - Plaintext: Uppercase letters string without punctuation and spaces.
- Output:
    - Ciphertext:  Hexadecimal string (and letters a-e in lower case).
- Preparation:
    - Key(SHA hash value)
        - Performing SHA-256 on "SECRET_KEY" results in '73ef2a4edd7a7fbf07fd5f6faf99674dc0c25a025fd74c221f4c35849e5c0fb3'.
- Encryption Steps:
    - Converts a plaintext string into a sequence of bytes (ASCII encoding).
    - Each byte is encrypted with the corresponding byte of the key using a bitwise different-or (XOR) operation. Repeatedly recycle the key so that the key length is the same as the number of plaintext bytes.
    - The encrypted byte sequence is converted to a hexadecimal string for ciphertext output.

 **Decryption Rules:**

- Input:
    - Ciphertext: Hexadecimal string (and letters a-e in lower case).
- Output:
    - Plaintext: Uppercase letters string without punctuation and spaces.
- Preparation:
    - Key (Same as encryption,is a hashed hash obtained by SHA_256)
- Decryption Steps:
    - Converts the ciphertext string into a sequence of bytes.
    - Each byte is decrypted (the decryption process is the same as the encryption process) with the corresponding byte of the key using a bitwise different-or (XOR) operation.
    - The decrypted byte sequence is converted to a plaintext string.
encoded text: 3da0640d88362ffe45b41326fbc0 decrypt to original information: nonculpability
encoded text: 3da0640d88362ffe45b41326fbc0 decrypt to original information: nonculpability
encoded text: 3da0640d88362ffe45b41326fbc0 decrypt to original information: nonculpability
encoded text: 23b6780789333c decrypt to original information: ? 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
```

Output:
开始解密过程...
加密的十六进制文本: 23b6780789333c
生成的SHA-256密钥: 73ef2a4edd7a7fbf07fd5f6faf99674dc0c25a025fd74c221f4c35849e5c0fb3
十六进制转换为字节序列: b'#\xb6x\x07\x893<'
开始XOR解密...
解密后的字节序列: b'PYRITIC'
最终解密结果: PYRITIC


Prompt:
**Encryption Rules:**

- Input:
    - Plaintext: Uppercase letters string without punctuation and spaces.
- Output:
    - Ciphertext:  A string without punctuation and spaces.
- Preparation:
    - Maximum number of characters per line: 5
- Encryption Steps:
    - Lines are counted from 1.
    - The plaintext is arranged in a special way: an odd number of lines are written from left to right, and an even number of lines are written from right to left, with a maximum of five letters per line.
        - For example, for the plaintext "LIDAHELLOWORLD", first write the first line, from left to right as LIDAH, and then write the second line, from right to left as WOLLE, and then write the third line, from left to right as ORLD, the entire content of the finished writing is expressed as follows
        LIDAH
        WOLLE
        ORLD
    - Then read by column, each column is read from the top down, each read a column plus a "#", read out the contents of the final ciphertext.
        - For example, the above written content is read as LWO#IOR#DLL#ALD#HE# by column.

plain text: nonculpability encode to encrypted text: NIL#OBI#NAT#CPY#UL#
plain text: nonculpability encode to encrypted text: NIL#OBI#NAT#CPY#UL#
plain text: nervii encode to encrypted 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: NERVII

Output:
1. 输入文本转换为大写并去除非字母字符: NERVII
2. 确定网格大小: 2行 x 5列
3. 开始填充网格:
   第1行从左到右填充: N E R V I
   第2行从右到左填充: I
4. 按列读取并添加#号:
   第1列: N##
   第2列: E##
   第3列: R##
   第4列: V##
   第5列: II#
5. 最终密文: N#E#R#V#II#


Prompt:
Please encode the plaintext step by step, ensuring the process is detailed and rigorous, and the result is correct.
**Encryption Rules:**

- Input:
    - Plaintext: Uppercase letters string without punctuation and spaces.
- Output:
    - Ciphertext:  A string without punctuation and spaces.
- Preparation:
    - Maximum number of characters per line: 5
- Encryption Steps:
    - Lines are counted from 1.
    - The plaintext is arranged in a special way: an odd number of lines are written from left to right, and an even number of lines are written from right to left, with a maximum of five letters per line.
        - For example, for the plaintext "LIDAHELLOWORLD", first write the first line, from left to right as LIDAH, and then write the second line, from right to left as WOLLE, and then write the third line, from left to right as ORLD, the entire content of the finished writing is expressed as follows
        LIDAH
        WOLLE
        ORLD
    - Then read by column, each column is read from the top down, each read a column plus a "#", read out the contents of the final ciphertext.
        - For example, the above written content is read as LWO#IOR#DLL#ALD#HE# by column.

plain text: nonculpability encode into encrypted text is: ?  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:
1. 输入文本转换为大写并去除非字母字符: NONCULPABILITY
2. 确定网格大小: 3行 x 5列
3. 开始填充网格:
   第1行从左到右填充: N O N C U
   第2行从右到左填充: L P A B I
   第3行从左到右填充: L I T Y
4. 按列读取并添加#号:
   第1列: NIL#
   第2列: OBI#
   第3列: NAT#
   第4列: CPY#
   第5列: UL##
5. 最终密文: NIL#OBI#NAT#CPY#UL#


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:  A string without punctuation and spaces.
- Preparation:
    - Maximum number of characters per line: 5
- Encryption Steps:
    - Lines are counted from 1.
    - The plaintext is arranged in a special way: an odd number of lines are written from left to right, and an even number of lines are written from right to left, with a maximum of five letters per line.
        - For example, for the plaintext "LIDAHELLOWORLD", first write the first line, from left to right as LIDAH, and then write the second line, from right to left as WOLLE, and then write the third line, from left to right as ORLD, the entire content of the finished writing is expressed as follows
        LIDAH
        WOLLE
        ORLD
    - Then read by column, each column is read from the top down, each read a column plus a "#", read out the contents of the final ciphertext.
        - For example, the above written content is read as LWO#IOR#DLL#ALD#HE# by column.

plain text: nervii encode into encoded text: N#E#R#V#II#
plain text: pyritic encode into encoded text: ?

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

Input: PYRITIC

Output:
1. 输入文本转换为大写并去除非字母字符: PYRITIC
2. 确定网格大小: 2行 x 5列
3. 开始填充网格:
   第1行从左到右填充: P Y R I T
   第2行从右到左填充: I C
4. 按列读取并添加#号:
   第1列: P##
   第2列: Y##
   第3列: R##
   第4列: IC#
   第5列: TI#
5. 最终密文: P#Y#R#IC#TI#


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:
    - 非加密信息: 仅包含大写字母的字符串，不含标点和空格
- 输出:
    - 暗文: 大写字母字符串
- 准备:
    - 标准字母表: "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
    - 反转字母表: "ZYXWVUTSRQPONMLKJIHGFEDCBA"
    - 替换字母表: "RFDJUHABCEGIKLMNOPQSTVWXYZ"
- 加密步骤:
    - 对非加密信息中的每个字母p:
    - (1) 使用反转字母表进行反向映射。在标准字母表中找到其位置，并用反转字母表中对应位置的字母替换。例如，A映射为Z，B映射为Y。
    - (2) 将步骤(1)得到的字母在标准字母表顺序中向前移动4位。例如，如果p=A，经过步骤(1)映射为Z，然后Z在标准字母表中向前移动4位得到D。
    - (3) 将步骤(2)得到的字母，在标准字母表中找到其位置，用替换字母表中对应位置的字母替换，得到最终的暗文字母。例如，如果经过步骤(2)得到字母D，则映射为J。
解密算法:
解密规则:
- 输入:
    - 暗文: 大写字母字符串
- 输出:
    - 非加密信息: 大写字母字符串
- 准备:
    - 标准字母表: "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
    - 反转字母表: "ZYXWVUTSRQPONMLKJIHGFEDCBA"
    - 替换字母表: "RFDJUHABCEGIKLMNOPQSTVWXYZ"
- 解密步骤(与加密步骤完全相反):
    - (1) 对暗文中的每个字母c，在替换字母表中找到其位置，用标准字母表中对应位置的字母替换。
    - (2) 将步骤(1)得到的字母按标准字母表顺序向后移动4位。
    - (3) 将步骤(2)得到的字母，在反转字母表中找到其位置，然后用标准字母表中对应位置的字母替换。例如，Z映射为A，Y映射为B。
暗文: ONOFEQMJDVQVGH 解密成非加密信息: nonculpability
暗文: OZKCVV 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: OZKCVV

Output:
标准字母表: ABCDEFGHIJKLMNOPQRSTUVWXYZ
反转字母表: ZYXWVUTSRQPONMLKJIHGFEDCBA
替换字母表: RFDJUHABCEGIKLMNOPQSTVWXYZ

解密字符 O:
1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: O -> Q
2. 将得到的字母向后移动4位: Q -> M
3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: M -> N

解密字符 Z:
1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: Z -> Z
2. 将得到的字母向后移动4位: Z -> V
3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: V -> E

解密字符 K:
1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: K -> M
2. 将得到的字母向后移动4位: M -> I
3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: I -> R

解密字符 C:
1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: C -> I
2. 将得到的字母向后移动4位: I -> E
3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: E -> V

解密字符 V:
1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: V -> V
2. 将得到的字母向后移动4位: V -> R
3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: R -> I

解密字符 V:
1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: V -> V
2. 将得到的字母向后移动4位: V -> R
3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: R -> I

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:
    - 初始文本: 仅包含大写字母的字符串，不含标点和空格
- 输出:
    - 密文: 大写字母字符串
- 准备:
    - 标准字母表: "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
    - 反转字母表: "ZYXWVUTSRQPONMLKJIHGFEDCBA"
    - 替换字母表: "RFDJUHABCEGIKLMNOPQSTVWXYZ"
- 加密步骤:
    - 对初始文本中的每个字母p:
    - (1) 使用反转字母表进行反向映射。在标准字母表中找到其位置，并用反转字母表中对应位置的字母替换。例如，A映射为Z，B映射为Y。
    - (2) 将步骤(1)得到的字母在标准字母表顺序中向前移动4位。例如，如果p=A，经过步骤(1)映射为Z，然后Z在标准字母表中向前移动4位得到D。
    - (3) 将步骤(2)得到的字母，在标准字母表中找到其位置，用替换字母表中对应位置的字母替换，得到最终的密文字母。例如，如果经过步骤(2)得到字母D，则映射为J。
解密算法:
解密规则:
- 输入:
    - 密文: 大写字母字符串
- 输出:
    - 初始文本: 大写字母字符串
- 准备:
    - 标准字母表: "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
    - 反转字母表: "ZYXWVUTSRQPONMLKJIHGFEDCBA"
    - 替换字母表: "RFDJUHABCEGIKLMNOPQSTVWXYZ"
- 解密步骤(与加密步骤完全相反):
    - (1) 对密文中的每个字母c，在替换字母表中找到其位置，用标准字母表中对应位置的字母替换。
    - (2) 将步骤(1)得到的字母按标准字母表顺序向后移动4位。
    - (3) 将步骤(2)得到的字母，在反转字母表中找到其位置，然后用标准字母表中对应位置的字母替换。例如，Z映射为A，Y映射为B。
密文: MHKVGVF 解码为初始文本: pyritic
密文: ONOFEQMJDVQVGH 解码为初始文本: ? 请一步一步完成，确保过程详细严谨，结果正确。

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

Input: ONOFEQMJDVQVGH

Output:
标准字母表: ABCDEFGHIJKLMNOPQRSTUVWXYZ
反转字母表: ZYXWVUTSRQPONMLKJIHGFEDCBA
替换字母表: RFDJUHABCEGIKLMNOPQSTVWXYZ

解密字符 O:
1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: O -> Q
2. 将得到的字母向后移动4位: Q -> M
3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: M -> N

解密字符 N:
1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: N -> P
2. 将得到的字母向后移动4位: P -> L
3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: L -> O

解密字符 O:
1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: O -> Q
2. 将得到的字母向后移动4位: Q -> M
3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: M -> N

解密字符 F:
1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: F -> B
2. 将得到的字母向后移动4位: B -> X
3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: X -> C

解密字符 E:
1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: E -> J
2. 将得到的字母向后移动4位: J -> F
3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: F -> U

解密字符 Q:
1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: Q -> S
2. 将得到的字母向后移动4位: S -> O
3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: O -> L

解密字符 M:
1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: M -> O
2. 将得到的字母向后移动4位: O -> K
3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: K -> P

解密字符 J:
1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: J -> D
2. 将得到的字母向后移动4位: D -> Z
3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: Z -> A

解密字符 D:
1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: D -> C
2. 将得到的字母向后移动4位: C -> Y
3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: Y -> B

解密字符 V:
1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: V -> V
2. 将得到的字母向后移动4位: V -> R
3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: R -> I

解密字符 Q:
1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: Q -> S
2. 将得到的字母向后移动4位: S -> O
3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: O -> L

解密字符 V:
1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: V -> V
2. 将得到的字母向后移动4位: V -> R
3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: R -> I

解密字符 G:
1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: G -> K
2. 将得到的字母向后移动4位: K -> G
3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: G -> T

解密字符 H:
1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: H -> F
2. 将得到的字母向后移动4位: F -> B
3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: B -> Y

最终解密结果: 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:
- 输入:
    - 原始信息: 仅包含大写字母的字符串，不含标点和空格
- 输出:
    - 加密文本: 大写字母字符串
- 准备:
    - 标准字母表: "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
    - 反转字母表: "ZYXWVUTSRQPONMLKJIHGFEDCBA"
    - 替换字母表: "RFDJUHABCEGIKLMNOPQSTVWXYZ"
- 加密步骤:
    - 对原始信息中的每个字母p:
    - (1) 使用反转字母表进行反向映射。在标准字母表中找到其位置，并用反转字母表中对应位置的字母替换。例如，A映射为Z，B映射为Y。
    - (2) 将步骤(1)得到的字母在标准字母表顺序中向前移动4位。例如，如果p=A，经过步骤(1)映射为Z，然后Z在标准字母表中向前移动4位得到D。
    - (3) 将步骤(2)得到的字母，在标准字母表中找到其位置，用替换字母表中对应位置的字母替换，得到最终的加密文本字母。例如，如果经过步骤(2)得到字母D，则映射为J。
解密算法:
解密规则:
- 输入:
    - 加密文本: 大写字母字符串
- 输出:
    - 原始信息: 大写字母字符串
- 准备:
    - 标准字母表: "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
    - 反转字母表: "ZYXWVUTSRQPONMLKJIHGFEDCBA"
    - 替换字母表: "RFDJUHABCEGIKLMNOPQSTVWXYZ"
- 解密步骤(与加密步骤完全相反):
    - (1) 对加密文本中的每个字母c，在替换字母表中找到其位置，用标准字母表中对应位置的字母替换。
    - (2) 将步骤(1)得到的字母按标准字母表顺序向后移动4位。
    - (3) 将步骤(2)得到的字母，在反转字母表中找到其位置，然后用标准字母表中对应位置的字母替换。例如，Z映射为A，Y映射为B。
加密文本: OZKCVV 解密成为原始信息: nervii
加密文本: MHKVGVF 解密成为原始信息: ? 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: MHKVGVF

Output:
标准字母表: ABCDEFGHIJKLMNOPQRSTUVWXYZ
反转字母表: ZYXWVUTSRQPONMLKJIHGFEDCBA
替换字母表: RFDJUHABCEGIKLMNOPQSTVWXYZ

解密字符 M:
1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: M -> O
2. 将得到的字母向后移动4位: O -> K
3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: K -> P

解密字符 H:
1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: H -> F
2. 将得到的字母向后移动4位: F -> B
3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: B -> Y

解密字符 K:
1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: K -> M
2. 将得到的字母向后移动4位: M -> I
3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: I -> R

解密字符 V:
1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: V -> V
2. 将得到的字母向后移动4位: V -> R
3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: R -> I

解密字符 G:
1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: G -> K
2. 将得到的字母向后移动4位: K -> G
3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: G -> T

解密字符 V:
1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: V -> V
2. 将得到的字母向后移动4位: V -> R
3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: R -> I

解密字符 F:
1. 在替换字母表中找到位置并用标准字母表对应位置字母替换: F -> B
2. 将得到的字母向后移动4位: B -> X
3. 在反转字母表中找到位置并用标准字母表对应位置字母替换: X -> C

最终解密结果: PYRITIC


Prompt:
Please encode the plaintext step by step, ensuring the process is detailed and rigorous, and the result is correct.
**Encryption Rules:**

- Input:
    - Plaintext: Uppercase letters string without punctuation and spaces.
- Output:
    - Ciphertext: Uppercase letters string.
- Preparation:
    - Reels

        Reel1:   "ABCEIGDJFVUYMHTQKZOLRXSPWN",
        Reel2:   "ACDEHFIJKTLMOUVYGZNPQXRWSB",
        Reel3:   "ADKOMJUBGEPHSCZINXFYQRTVWL",
        Reel4:   "AEDCBIFGJHLKMRUOQVPTNWYXZS",
        Reel5:   "AFNQUKDOPITJBRHCYSLWEMZVXG",
        Reel6:   "AGPOCIXLURNDYZHWBJSQFKVMET",
        Reel7:    "AHXJEZBNIKPVROGSYDULCFMQTW",
        Reel8:    "AIHPJOBWKCVFZLQERYNSUMGTDX",
        Reel9:    "AJDSKQOIVTZEFHGYUNLPMBXWCR",
        Reel10:   "AKELBDFJGHONMTPRQSVZUXYWIC",
        Reel11:    "ALTMSXVQPNOHUWDIZYCGKRFBEJ",
        Reel12:    "AMNFLHQGCUJTBYPZKXISRDVEWO",
        Reel13:    "ANCJILDHBMKGXUZTSWQYVORPFE",
        Reel14:    "AODWPKJVIUQHZCTXBLEGNYRSMF",
        Reel15:    "APBVHIYKSGUENTCXOWFQDRLJZM",
        Reel16:    "AQJNUBTGIMWZRVLXCSHDEOKFPY",
        Reel17:    "ARMYOFTHEUSZJXDPCWGQIBKLNV",
        Reel18:    "ASDMCNEQBOZPLGVJRKYTFUIWXH",
        Reel19:    "ATOJYLFXNGWHVCMIRBSEKUPDZQ",
        Reel20:    "AUTRZXQLYIOVBPESNHJWMDGFCK",
        Reel21:     "AVNKHRGOXEYBFSJMUDQCLZWTIP",
        Reel22:    "AWVSFDLIEBHKNRJQZGMXPUCOTY",
        Reel23:    "AXKWREVDTUFOYHMLSIQNJCPGBZ",
        Reel24:    "AYJPXMVKBQWUGLOSTECHNZFRID",
        Reel25:    "AZDNBUHYFWJLVGRCQMPSOEXTKI"

- Encryption Steps:
    - Initially select the Reel1.
    - For each character p in the plaintext:
        - Find character p on the Reel and replace it with the next character on the Reel to get the ciphertext character.
        - If the current character is at the end of the Reel, go around to the beginning of the Reel.
        - Move to the next Reel to get the next character. When you reach the last Reel, circle back to the first Reel and continue the encryption process.

clear text: pyritic encode into encoded text: WGTFJXF
clear text: nonculpability encode into encoded text: AUXBKUVIXCTSSR
clear text: pyritic encode into encoded text: WGTFJXF
clear text: nervii encode into encoded 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: NERVII

Output:
处理后的输入文本: NERVII
初始化Jefferson密码轮...
开始加密过程:
- 从第1个密码轮开始
- 对每个字符:
  * 在第1个密码轮上，将字符 N 替换为下一个字符 A
  * 在第2个密码轮上，将字符 E 替换为下一个字符 H
  * 在第3个密码轮上，将字符 R 替换为下一个字符 T
  * 在第4个密码轮上，将字符 V 替换为下一个字符 P
  * 在第5个密码轮上，将字符 I 替换为下一个字符 T
  * 在第6个密码轮上，将字符 I 替换为下一个字符 X
加密完成，结果: AHTPTX


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: Uppercase letters string.
- Preparation:
    - Reels

        Reel1:   "ABCEIGDJFVUYMHTQKZOLRXSPWN",
        Reel2:   "ACDEHFIJKTLMOUVYGZNPQXRWSB",
        Reel3:   "ADKOMJUBGEPHSCZINXFYQRTVWL",
        Reel4:   "AEDCBIFGJHLKMRUOQVPTNWYXZS",
        Reel5:   "AFNQUKDOPITJBRHCYSLWEMZVXG",
        Reel6:   "AGPOCIXLURNDYZHWBJSQFKVMET",
        Reel7:    "AHXJEZBNIKPVROGSYDULCFMQTW",
        Reel8:    "AIHPJOBWKCVFZLQERYNSUMGTDX",
        Reel9:    "AJDSKQOIVTZEFHGYUNLPMBXWCR",
        Reel10:   "AKELBDFJGHONMTPRQSVZUXYWIC",
        Reel11:    "ALTMSXVQPNOHUWDIZYCGKRFBEJ",
        Reel12:    "AMNFLHQGCUJTBYPZKXISRDVEWO",
        Reel13:    "ANCJILDHBMKGXUZTSWQYVORPFE",
        Reel14:    "AODWPKJVIUQHZCTXBLEGNYRSMF",
        Reel15:    "APBVHIYKSGUENTCXOWFQDRLJZM",
        Reel16:    "AQJNUBTGIMWZRVLXCSHDEOKFPY",
        Reel17:    "ARMYOFTHEUSZJXDPCWGQIBKLNV",
        Reel18:    "ASDMCNEQBOZPLGVJRKYTFUIWXH",
        Reel19:    "ATOJYLFXNGWHVCMIRBSEKUPDZQ",
        Reel20:    "AUTRZXQLYIOVBPESNHJWMDGFCK",
        Reel21:     "AVNKHRGOXEYBFSJMUDQCLZWTIP",
        Reel22:    "AWVSFDLIEBHKNRJQZGMXPUCOTY",
        Reel23:    "AXKWREVDTUFOYHMLSIQNJCPGBZ",
        Reel24:    "AYJPXMVKBQWUGLOSTECHNZFRID",
        Reel25:    "AZDNBUHYFWJLVGRCQMPSOEXTKI"

- Encryption Steps:
    - Initially select the Reel1.
    - For each character p in the plaintext:
        - Find character p on the Reel and replace it with the next character on the Reel to get the ciphertext character.
        - If the current character is at the end of the Reel, go around to the beginning of the Reel.
        - Move to the next Reel to get the next character. When you reach the last Reel, circle back to the first Reel and continue the encryption process.

original information: nervii encrypt into encoded text: AHTPTX
original information: pyritic encrypt into encoded text: WGTFJXF
original information: nonculpability 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:
处理后的输入文本: NONCULPABILITY
初始化Jefferson密码轮...
开始加密过程:
- 从第1个密码轮开始
- 对每个字符:
  * 在第1个密码轮上，将字符 N 替换为下一个字符 A
  * 在第2个密码轮上，将字符 O 替换为下一个字符 U
  * 在第3个密码轮上，将字符 N 替换为下一个字符 X
  * 在第4个密码轮上，将字符 C 替换为下一个字符 B
  * 在第5个密码轮上，将字符 U 替换为下一个字符 K
  * 在第6个密码轮上，将字符 L 替换为下一个字符 U
  * 在第7个密码轮上，将字符 P 替换为下一个字符 V
  * 在第8个密码轮上，将字符 A 替换为下一个字符 I
  * 在第9个密码轮上，将字符 B 替换为下一个字符 X
  * 在第10个密码轮上，将字符 I 替换为下一个字符 C
  * 在第11个密码轮上，将字符 L 替换为下一个字符 T
  * 在第12个密码轮上，将字符 I 替换为下一个字符 S
  * 在第13个密码轮上，将字符 T 替换为下一个字符 S
  * 在第14个密码轮上，将字符 Y 替换为下一个字符 R
加密完成，结果: AUXBKUVIXCTSSR


Prompt:
Your task is to convert the plaintext into an unreadable form usingthe appropriate algorithm, ensuring the security of the information during transmission.
**Encryption Rules:**

- Input:
    - Plaintext: Uppercase letters string without punctuation and spaces.
- Output:
    - Ciphertext: Uppercase letters string.
- Preparation:
    - Reels

        Reel1:   "ABCEIGDJFVUYMHTQKZOLRXSPWN",
        Reel2:   "ACDEHFIJKTLMOUVYGZNPQXRWSB",
        Reel3:   "ADKOMJUBGEPHSCZINXFYQRTVWL",
        Reel4:   "AEDCBIFGJHLKMRUOQVPTNWYXZS",
        Reel5:   "AFNQUKDOPITJBRHCYSLWEMZVXG",
        Reel6:   "AGPOCIXLURNDYZHWBJSQFKVMET",
        Reel7:    "AHXJEZBNIKPVROGSYDULCFMQTW",
        Reel8:    "AIHPJOBWKCVFZLQERYNSUMGTDX",
        Reel9:    "AJDSKQOIVTZEFHGYUNLPMBXWCR",
        Reel10:   "AKELBDFJGHONMTPRQSVZUXYWIC",
        Reel11:    "ALTMSXVQPNOHUWDIZYCGKRFBEJ",
        Reel12:    "AMNFLHQGCUJTBYPZKXISRDVEWO",
        Reel13:    "ANCJILDHBMKGXUZTSWQYVORPFE",
        Reel14:    "AODWPKJVIUQHZCTXBLEGNYRSMF",
        Reel15:    "APBVHIYKSGUENTCXOWFQDRLJZM",
        Reel16:    "AQJNUBTGIMWZRVLXCSHDEOKFPY",
        Reel17:    "ARMYOFTHEUSZJXDPCWGQIBKLNV",
        Reel18:    "ASDMCNEQBOZPLGVJRKYTFUIWXH",
        Reel19:    "ATOJYLFXNGWHVCMIRBSEKUPDZQ",
        Reel20:    "AUTRZXQLYIOVBPESNHJWMDGFCK",
        Reel21:     "AVNKHRGOXEYBFSJMUDQCLZWTIP",
        Reel22:    "AWVSFDLIEBHKNRJQZGMXPUCOTY",
        Reel23:    "AXKWREVDTUFOYHMLSIQNJCPGBZ",
        Reel24:    "AYJPXMVKBQWUGLOSTECHNZFRID",
        Reel25:    "AZDNBUHYFWJLVGRCQMPSOEXTKI"

- Encryption Steps:
    - Initially select the Reel1.
    - For each character p in the plaintext:
        - Find character p on the Reel and replace it with the next character on the Reel to get the ciphertext character.
        - If the current character is at the end of the Reel, go around to the beginning of the Reel.
        - Move to the next Reel to get the next character. When you reach the last Reel, circle back to the first Reel and continue the encryption process.

original information: nonculpability encrypt to encoded text: AUXBKUVIXCTSSR
original information: nonculpability encrypt to encoded text: AUXBKUVIXCTSSR
original information: nervii encrypt to encoded text: AHTPTX
original information: pyritic encrypt to encoded text is: ?  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: PYRITIC

Output:
处理后的输入文本: PYRITIC
初始化Jefferson密码轮...
开始加密过程:
- 从第1个密码轮开始
- 对每个字符:
  * 在第1个密码轮上，将字符 P 替换为下一个字符 W
  * 在第2个密码轮上，将字符 Y 替换为下一个字符 G
  * 在第3个密码轮上，将字符 R 替换为下一个字符 T
  * 在第4个密码轮上，将字符 I 替换为下一个字符 F
  * 在第5个密码轮上，将字符 T 替换为下一个字符 J
  * 在第6个密码轮上，将字符 I 替换为下一个字符 X
  * 在第7个密码轮上，将字符 C 替换为下一个字符 F
加密完成，结果: WGTFJXF


Prompt:
As an encoder, your task is to use the encryption algorithm described in the examples to encrypt the plaintext.
plain text: nonculpability encrypt into cipher text: [110, 111, 110, 99, 117, 108, 112, 97, 98, 105, 108, 105, 116, 121]
plain text: nonculpability encrypt into cipher text: [110, 111, 110, 99, 117, 108, 112, 97, 98, 105, 108, 105, 116, 121]
plain text: pyritic encrypt into cipher text: [112, 121, 114, 105, 116, 105, 99]
plain text: nervii encrypt into cipher text is: ?  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: nervii

Output:
Character:   n, ASCII Code: 110
Character:   e, ASCII Code: 101
Character:   r, ASCII Code: 114
Character:   v, ASCII Code: 118
Character:   i, ASCII Code: 105
Character:   i, ASCII Code: 105
ASCII List: [110, 101, 114, 118, 105, 105]


Prompt:
Your task is to convert the plaintext into an unreadable form usingthe appropriate algorithm, ensuring the security of the information during transmission.
plain text: nervii encrypt to encrypted text: [110, 101, 114, 118, 105, 105]
plain text: nervii encrypt to encrypted text: [110, 101, 114, 118, 105, 105]
plain text: nonculpability encrypt to encrypted 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: nonculpability

Output:
Character:   n, ASCII Code: 110
Character:   o, ASCII Code: 111
Character:   n, ASCII Code: 110
Character:   c, ASCII Code: 99
Character:   u, ASCII Code: 117
Character:   l, ASCII Code: 108
Character:   p, ASCII Code: 112
Character:   a, ASCII Code: 97
Character:   b, ASCII Code: 98
Character:   i, ASCII Code: 105
Character:   l, ASCII Code: 108
Character:   i, ASCII Code: 105
Character:   t, ASCII Code: 116
Character:   y, ASCII Code: 121
ASCII List: [110, 111, 110, 99, 117, 108, 112, 97, 98, 105, 108, 105, 116, 121]


Prompt:
Your task is to convert the plaintext into an unreadable form usingthe appropriate algorithm, ensuring the security of the information during transmission.
clear text: nervii encode to cipher text: [110, 101, 114, 118, 105, 105]
clear text: nervii encode to cipher text: [110, 101, 114, 118, 105, 105]
clear text: nonculpability encode to cipher text: [110, 111, 110, 99, 117, 108, 112, 97, 98, 105, 108, 105, 116, 121]
clear text: nervii encode to cipher text: [110, 101, 114, 118, 105, 105]
clear text: pyritic encode 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: pyritic

Output:
Character:   p, ASCII Code: 112
Character:   y, ASCII Code: 121
Character:   r, ASCII Code: 114
Character:   i, ASCII Code: 105
Character:   t, ASCII Code: 116
Character:   i, ASCII Code: 105
Character:   c, ASCII Code: 99
ASCII List: [112, 121, 114, 105, 116, 105, 99]


Prompt:
As a decoder, your task is to use the encryption algorithm described in the examples to decrypt the ciphertext.
Please decrypt the password according to the encryption and decryption algorithms.

Encryption Algorithm:
The Atbash cipher encoding rule is to replace each letter with its reverse corresponding letter in the alphabet. For example, 'A' becomes 'Z', 'B' becomes 'Y', and so on. Non-letter characters remain unchanged.

Decryption Algorithm:
The Atbash cipher decoding rule is to replace each letter with its reverse corresponding letter in the alphabet. Since the Atbash cipher is symmetric, the decoding process is the same as the encoding process. For example, 'Z' becomes 'A', 'Y' becomes 'B', and so on. Non-letter characters remain unchanged.
cipher text: mlmxfokzyrorgb decrypt to plain text: nonculpability
cipher text: kbirgrx decrypt to plain text: pyritic
cipher text: kbirgrx decrypt to plain text: pyritic
cipher text: mvierr decrypt to plain text is: ?

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

Input: mvierr

Output:
解码过程:
转换单词: mvierr
  'm' -> 'n'
  'v' -> 'e'
  'i' -> 'r'
  'e' -> 'v'
  'r' -> 'i'
  'r' -> 'i'
解码后的文本: nervii


Prompt:
Please decode the ciphertext according to the examples and the given information.
Please decrypt the password according to the encryption and decryption algorithms.

Encryption Algorithm:
The Atbash cipher encoding rule is to replace each letter with its reverse corresponding letter in the alphabet. For example, 'A' becomes 'Z', 'B' becomes 'Y', and so on. Non-letter characters remain unchanged.

Decryption Algorithm:
The Atbash cipher decoding rule is to replace each letter with its reverse corresponding letter in the alphabet. Since the Atbash cipher is symmetric, the decoding process is the same as the encoding process. For example, 'Z' becomes 'A', 'Y' becomes 'B', and so on. Non-letter characters remain unchanged.
cipher text: mvierr decrypt into clear text: nervii
cipher text: kbirgrx decrypt into clear text: pyritic
cipher text: mlmxfokzyrorgb decrypt into clear 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: mlmxfokzyrorgb

Output:
解码过程:
转换单词: mlmxfokzyrorgb
  'm' -> 'n'
  'l' -> 'o'
  'm' -> 'n'
  'x' -> 'c'
  'f' -> 'u'
  'o' -> 'l'
  'k' -> 'p'
  'z' -> 'a'
  'y' -> 'b'
  'r' -> 'i'
  'o' -> 'l'
  'r' -> 'i'
  'g' -> 't'
  'b' -> 'y'
解码后的文本: nonculpability


Prompt:
Please decrypt the password according to the encryption and decryption algorithms.

Encryption Algorithm:
The Atbash cipher encoding rule is to replace each letter with its reverse corresponding letter in the alphabet. For example, 'A' becomes 'Z', 'B' becomes 'Y', and so on. Non-letter characters remain unchanged.

Decryption Algorithm:
The Atbash cipher decoding rule is to replace each letter with its reverse corresponding letter in the alphabet. Since the Atbash cipher is symmetric, the decoding process is the same as the encoding process. For example, 'Z' becomes 'A', 'Y' becomes 'B', and so on. Non-letter characters remain unchanged.
encoded text: kbirgrx decrypt to plain text is: ?

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

Input: kbirgrx

Output:
解码过程:
转换单词: kbirgrx
  'k' -> 'p'
  'b' -> 'y'
  'i' -> 'r'
  'r' -> 'i'
  'g' -> 't'
  'r' -> 'i'
  'x' -> 'c'
解码后的文本: pyritic