You are given a m x n matrix mat of positive integers.
Return an integer denoting the number of ways to choose exactly one integer from each row of mat such that the greatest common divisor of all chosen integers is 1.
Since the answer may be very large, return it modulo 109 + 7.
\n
Example 1:
\n\nInput: mat = [[1,2],[3,4]]
\n\nOutput: 3
\n\nExplanation:
\n\n| Chosen integer in the first row | \n\t\t\tChosen integer in the second row | \n\t\t\tGreatest common divisor of chosen integers | \n\t\t
|---|---|---|
| 1 | \n\t\t\t3 | \n\t\t\t1 | \n\t\t
| 1 | \n\t\t\t4 | \n\t\t\t1 | \n\t\t
| 2 | \n\t\t\t3 | \n\t\t\t1 | \n\t\t
| 2 | \n\t\t\t4 | \n\t\t\t2 | \n\t\t
3 of these combinations have a greatest common divisor of 1. Therefore, the answer is 3.
\nExample 2:
\n\nInput: mat = [[2,2],[2,2]]
\n\nOutput: 0
\n\nExplanation:
\n\nEvery combination has a greatest common divisor of 2. Therefore, the answer is 0.
\n\n
Constraints:
\n\n1 <= m == mat.length <= 1501 <= n == mat[i].length <= 1501 <= mat[i][j] <= 150给你一个由正整数组成的 m x n 矩阵 mat。
返回一个整数，表示从 mat 的每一行中 恰好 选择一个整数，使得所有被选整数的 最大公约数 为 1 的选择方案数量。
由于答案可能非常大，请将其 模 109 + 7 后返回。
\n\n
示例 1:
\n\n输入: mat = [[1,2],[3,4]]
\n\n输出: 3
\n\n解释:
\n\n| 第一行中选择的整数 | \n\t\t\t第二行中选择的整数 | \n\t\t\t被选整数的最大公约数 | \n\t\t
|---|---|---|
| 1 | \n\t\t\t3 | \n\t\t\t1 | \n\t\t
| 1 | \n\t\t\t4 | \n\t\t\t1 | \n\t\t
| 2 | \n\t\t\t3 | \n\t\t\t1 | \n\t\t
| 2 | \n\t\t\t4 | \n\t\t\t2 | \n\t\t
其中 3 种组合的最大公约数为 1。因此，答案是 3。
\n示例 2:
\n\n输入: mat = [[2,2],[2,2]]
\n\n输出: 0
\n\n解释:
\n\n所有组合的最大公约数都是 2。因此，答案是 0。
\n\n\n
提示:
\n\n1 <= m == mat.length <= 1501 <= n == mat[i].length <= 1501 <= mat[i][j] <= 150dp[row][g], where row is the current row and g is the current gcd value.",
"Initialize the first row: for each value v in row 0 do dp[0][v] += 1.",
"For a row i, use the values from the previous row i - 1 to build the values: for each previous gcd g and each value v in row i.",
"The final answer is dp[n-1][1] (number of ways with gcd 1)."
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