<p>You are given a <strong>0-indexed</strong> <code>m x n</code> integer matrix <code>grid</code> and an integer <code>k</code>. You are currently at position <code>(0, 0)</code> and you want to reach position <code>(m - 1, n - 1)</code> moving only <strong>down</strong> or <strong>right</strong>.</p>

<p>Return<em> the number of paths where the sum of the elements on the path is divisible by </em><code>k</code>. Since the answer may be very large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img src="https://assets.leetcode.com/uploads/2022/08/13/image-20220813183124-1.png" style="width: 437px; height: 200px;" />
<pre>
<strong>Input:</strong> grid = [[5,2,4],[3,0,5],[0,7,2]], k = 3
<strong>Output:</strong> 2
<strong>Explanation:</strong> There are two paths where the sum of the elements on the path is divisible by k.
The first path highlighted in red has a sum of 5 + 2 + 4 + 5 + 2 = 18 which is divisible by 3.
The second path highlighted in blue has a sum of 5 + 3 + 0 + 5 + 2 = 15 which is divisible by 3.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img src="https://assets.leetcode.com/uploads/2022/08/17/image-20220817112930-3.png" style="height: 85px; width: 132px;" />
<pre>
<strong>Input:</strong> grid = [[0,0]], k = 5
<strong>Output:</strong> 1
<strong>Explanation:</strong> The path highlighted in red has a sum of 0 + 0 = 0 which is divisible by 5.
</pre>

<p><strong class="example">Example 3:</strong></p>
<img src="https://assets.leetcode.com/uploads/2022/08/12/image-20220812224605-3.png" style="width: 257px; height: 200px;" />
<pre>
<strong>Input:</strong> grid = [[7,3,4,9],[2,3,6,2],[2,3,7,0]], k = 1
<strong>Output:</strong> 10
<strong>Explanation:</strong> Every integer is divisible by 1 so the sum of the elements on every possible path is divisible by k.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>m == grid.length</code></li>
	<li><code>n == grid[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 5 * 10<sup>4</sup></code></li>
	<li><code>1 &lt;= m * n &lt;= 5 * 10<sup>4</sup></code></li>
	<li><code>0 &lt;= grid[i][j] &lt;= 100</code></li>
	<li><code>1 &lt;= k &lt;= 50</code></li>
</ul>

class Solution:
    def numberOfPaths(self, grid: List[List[int]], k: int) -> int:
        m = len(grid)
        n = len(grid[0])
        MOD = 10 ** 9 + 7

        # @cache
        # def dfs(i, j, v):
        #     if i < 0 or j < 0: return 0
        #     v = (v + grid[i][j]) % k
        #     if i == j == 0: return v == 0
        #     return (dfs(i-1, j, v) + dfs(i, j-1, v)) % MOD
        # ans = dfs(m-1, n-1, 0)
        # dfs.cache_clear()
        # return int(ans)

        @cache
        def dfs(i, j, v):
            # if i >= m or j >= n: return 0
            v = (v + grid[i][j]) % k
            if i == m - 1 and j == n - 1:
                if v == 0:
                    return 1
                else:
                    return 0

            # Correct
            # if i >= m or j >= n: return 0
            # v = (v + grid[i][j]) % k
            # if i == m - 1 and j == n - 1: return v == 0
            # return (dfs(i+1, j, v) + dfs(i, j+1, v)) % MOD
            
            d = r = 0
            if i < m - 1:
                d = dfs(i+1, j, v)
            if j < n - 1:
                r = dfs(i, j + 1, v)
            return (d + r) % MOD
            # return (dfs(i+1, j, v) + dfs(i, j+1, v)) % MOD

        ans = dfs(0, 0, 0)
        dfs.cache_clear()
        return int(ans)
            


        #记忆化搜索不要把unhashable的data type当参数，就算换成tuple也还是会爆内存，cache clear也没用
        # @cache
        # def dfs(i, j, path):
        #     if i == m - 1 and j == n - 1:
        #         res = []
        #         for p in path:
        #             if (p + grid[i][j]) % k == 0:
        #                 res.append(1)
        #         return tuple(res)
            
        #     npath = tuple([(p + grid[i][j]) % k for p in path])
            
        #     down = right = tuple()
        #     if i < m - 1:
        #         down = dfs(i+1, j, npath)
        #     if j < n - 1:
        #         right = dfs(i, j+1, npath)

        #     return down + right

        # ans = len(dfs(0, 0, tuple([0])))
        # dfs.cache_clear()
        # return ans

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