<p>There is an <code>m x n</code> grid with a ball. The ball is initially at the position <code>[startRow, startColumn]</code>. You are allowed to move the ball to one of the four adjacent cells in the grid (possibly out of the grid crossing the grid boundary). You can apply <strong>at most</strong> <code>maxMove</code> moves to the ball.</p>

<p>Given the five integers <code>m</code>, <code>n</code>, <code>maxMove</code>, <code>startRow</code>, <code>startColumn</code>, return the number of paths to move the ball out of the grid boundary. Since the answer can 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 alt="" src="https://assets.leetcode.com/uploads/2021/04/28/out_of_boundary_paths_1.png" style="width: 500px; height: 296px;" />
<pre>
<strong>Input:</strong> m = 2, n = 2, maxMove = 2, startRow = 0, startColumn = 0
<strong>Output:</strong> 6
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/04/28/out_of_boundary_paths_2.png" style="width: 500px; height: 293px;" />
<pre>
<strong>Input:</strong> m = 1, n = 3, maxMove = 3, startRow = 0, startColumn = 1
<strong>Output:</strong> 12
</pre>

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

<ul>
	<li><code>1 &lt;= m, n &lt;= 50</code></li>
	<li><code>0 &lt;= maxMove &lt;= 50</code></li>
	<li><code>0 &lt;= startRow &lt; m</code></li>
	<li><code>0 &lt;= startColumn &lt; n</code></li>
</ul>

class Solution:
    def findPaths(self, m: int, n: int, maxMove: int, startRow: int, startColumn: int) -> int:

        dp = [[[1] * (maxMove + 1) for _ in range(n+1)] for _ in range(m+1)]

        @cache
        def numPaths(i,j,move):
            if i == -1 or i == m or j == -1 or j == n:
                return 1
            if move >= maxMove:
                return 0
            else:
                up = dp[i - 1][j][ move + 1]
                down = dp[i + 1][j][ move + 1]
                left = dp[i][j - 1] [move + 1]
                right = dp[i][j + 1] [move + 1]
                return up + down + left + right
        
        for k in range(maxMove, -1, -1):
            for i in range(0, m):
                for j in range(0, n):
                    if i == -1 or i == m or j == -1 or j == n:
                        dp[i][j][k] = 1
                    if k >= maxMove:
                        dp[i][j][k] = 0
                    else:
                        up = dp[i - 1][j][ k + 1]
                        down = dp[i + 1][j][ k + 1]
                        left = dp[i][j - 1] [k + 1]
                        print(i,j,k)
                        right = dp[i][j + 1] [k + 1]
                        dp[i][j][k] = up + down + left + right
        return dp[startRow] [startColumn] [0] % (10 ** 9 + 7)
            
21900