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"""#
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### 谜题描述
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You are given a matrix a consisting of positive integers. It has n rows and m columns.
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Construct a matrix b consisting of positive integers. It should have the same size as a, and the following conditions should be met:
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* 1 ≤ b_{i,j} ≤ 10^6;
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* b_{i,j} is a multiple of a_{i,j};
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* the absolute value of the difference between numbers in any adjacent pair of cells (two cells that share the same side) in b is equal to k^4 for some integer k ≥ 1 (k is not necessarily the same for all pairs, it is own for each pair).
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We can show that the answer always exists.
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Input
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The first line contains two integers n and m (2 ≤ n,m ≤ 500).
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Each of the following n lines contains m integers. The j-th integer in the i-th line is a_{i,j} (1 ≤ a_{i,j} ≤ 16).
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Output
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The output should contain n lines each containing m integers. The j-th integer in the i-th line should be b_{i,j}.
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Examples
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Input
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2 2
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1 2
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2 3
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Output
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1 2
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2 3
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Input
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2 3
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16 16 16
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16 16 16
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Output
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16 32 48
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32 48 64
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Input
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2 2
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3 11
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12 8
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Output
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327 583
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408 664
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Note
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In the first example, the matrix a can be used as the matrix b, because the absolute value of the difference between numbers in any adjacent pair of cells is 1 = 1^4.
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In the third example:
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* 327 is a multiple of 3, 583 is a multiple of 11, 408 is a multiple of 12, 664 is a multiple of 8;
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* |408 - 327| = 3^4, |583 - 327| = 4^4, |664 - 408| = 4^4, |664 - 583| = 3^4.
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Here is a reference code to solve this task. You can use this to help you genereate cases or validate the solution.
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```python
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from collections import Counter, defaultdict, deque
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import bisect
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from sys import stdin, stdout
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from itertools import repeat
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import math
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def inp(force_list=False):
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re = map(int, raw_input().split())
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if len(re) == 1 and not force_list:
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return re[0]
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return re
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def inst():
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return raw_input().strip()
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def gcd(x, y):
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while(y):
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x, y = y, x % y
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return x
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mod = int(1e9)+7
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def quickm(a, b):
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base = a
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re = 1
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while b:
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if b&1:
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re *= base
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re %= mod
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b >>= 1
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base *= base
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base %= mod
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return re
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def inv(num):
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return quickm(num, mod-2)
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def my_main():
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kase = 1 #inp()
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pans = []
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for _ in range(kase):
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n, m = inp()
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da = []
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for i in range(n):
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da.append(inp())
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for i in range(n):
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ans = []
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for j in range(m):
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if (i+j)%2:
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ans.append(720720)
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else:
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ans.append(720720+da[i][j]**4)
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pans.append(' '.join(map(str, ans)))
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print '\n'.join(pans)
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my_main()
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```
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请完成上述谜题的训练场环境类实现，包括所有必要的方法。
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"""
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from bootcamp import Basebootcamp
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import random
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from bootcamp import Basebootcamp
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class Dmultiplesandpowerdifferencesbootcamp(Basebootcamp):
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def __init__(self, n=2, m=2, min_a=1, max_a=16):
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self.n = n
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self.m = m
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self.min_a = min_a
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self.max_a = max_a
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def case_generator(self):
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a_matrix = [
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[random.randint(self.min_a, self.max_a) for _ in range(self.m)]
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for _ in range(self.n)
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]
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return {
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'n': self.n,
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'm': self.m,
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'a_matrix': a_matrix
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}
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@staticmethod
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def prompt_func(question_case):
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n = question_case['n']
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m = question_case['m']
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a_matrix = question_case['a_matrix']
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problem = (
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"你是一名数学谜题的解谜者，需要构造一个满足特定条件的矩阵。给定一个n行m列的矩阵a，其中每个元素都是正整数。你的任务是根据以下规则构造矩阵b：\n\n"
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"规则说明：\n"
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"1. 矩阵b的大小必须与a相同，并且每个元素b[i][j]必须是a[i][j]的正整数倍。\n"
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"2. 矩阵b中任意两个相邻元素（相邻指的是上下左右相邻，即共享同一条边的单元格）的绝对差必须是某个整数的四次方。例如，差可以是1⁴=1，2⁴=16，3⁴=81，依此类推。\n"
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"3. 每个元素b[i][j]的取值范围必须在1到1,000,000之间（包含1和1,000,000）。\n\n"
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"输入格式：\n"
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f"第一行包含两个整数n和m（此处n={n}，m={m}）。\n"
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"接下来的n行每行包含m个整数，表示矩阵a的元素。\n\n"
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"输出格式：\n"
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"输出n行，每行m个整数，表示满足条件的矩阵b。\n\n"
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"矩阵a的输入如下：\n"
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f"{n} {m}\n"
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)
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for row in a_matrix:
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problem += " ".join(map(str, row)) + "\n"
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problem += (
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"\n请按照输出格式要求，将正确的矩阵b输出，并将答案放置在[answer]标签内。例如：\n"
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"[answer]\n"
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"16 32\n"
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"48 64\n"
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"[/answer]"
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)
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return problem
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@staticmethod
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def extract_output(output):
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import re
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pattern = re.compile(r'\[answer\](.*?)\[/answer\]', re.DOTALL)
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matches = pattern.findall(output)
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if not matches:
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return None
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last_answer = matches[-1].strip()
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matrix = []
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for line in last_answer.split('\n'):
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line = line.strip()
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if not line:
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continue
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try:
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elements = list(map(int, line.split()))
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matrix.append(elements)
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except ValueError:
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return None
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if not matrix:
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return None
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cols = len(matrix[0])
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for row in matrix:
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if len(row) != cols:
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return None
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return matrix
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@classmethod
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def _verify_correction(cls, solution, identity):
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if not solution or not isinstance(solution, list) or not all(isinstance(row, list) for row in solution):
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return False
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n = identity['n']
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m = identity['m']
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a_matrix = identity['a_matrix']
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if len(solution) != n or any(len(row) != m for row in solution):
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return False
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for i in range(n):
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for j in range(m):
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b_val = solution[i][j]
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a_val = a_matrix[i][j]
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if not (1 <= b_val <= 10**6) or b_val % a_val != 0:
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return False
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for i in range(n):
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for j in range(m):
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current = solution[i][j]
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if j + 1 < m:
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right = solution[i][j+1]
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if not cls.is_k4(abs(current - right)):
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return False
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if i + 1 < n:
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down = solution[i+1][j]
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if not cls.is_k4(abs(current - down)):
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return False
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return True
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@staticmethod
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def is_k4(x):
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if x < 1:
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return False
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k = 1
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while True:
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k4 = k ** 4
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if k4 == x:
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return True
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if k4 > x:
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return False
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k += 1
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