<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:
        def tle_():
            COLS = len(grid[0])
            ROWS = len(grid)
            dp = [[list() for _ in range(COLS)] for r in range(ROWS)]
            dp[0][0] = [grid[0][0]]
            for c in range(1, COLS):
                dp[0][c].append(grid[0][c] + dp[0][c-1][0])
            for r in range(1,ROWS):
                dp[r][0].append(grid[r][0] + dp[r-1][0][0])

            for c in range(1,COLS):
                for r in range(1,ROWS):
                    above = dp[r-1][c]
                    left = dp[r][c-1]
                    if len(above) == 0:
                        dp[r][c].append(grid[r][c])
                    if len(left) == 0:
                        dp[r][c].append(grid[r][c])
                    for num in above:
                        dp[r][c].append(num + grid[r][c])
                    for num in left:
                        dp[r][c].append(num + grid[r][c])
            '''
            [[5,2,4]
            ,[3,0,5]
            ,[0,7,2]]
            '''
            count = 0
            for path in dp[ROWS-1][COLS-1]:
                if path % k == 0:
                    count += 1

            return count % (10**9 + 7)

        memo = {}
        MOD = 10**9 + 7
        def recurse(row, col, sumsofar):
            nonlocal MOD
            
            if row >= len(grid) or col >= len(grid[0]):
                return 0
            
            key = (row, col, sumsofar % k)
            if key in memo:
                return memo[key]
            if row == len(grid)-1 and col == len(grid[0]) - 1:
                sumsofar += grid[row][col]
                if (sumsofar % k) == 0:
                    return 1
                return 0

            down = recurse(row+1, col, sumsofar+grid[row][col]) % MOD
            right = recurse(row, col+1, sumsofar+grid[row][col]) % MOD
            memo[key] = (down + right) % MOD
            return memo[key]
        return recurse(0,0,0)

            



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