<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 {
    static const int MOD = 1e9 + 7;
    static constexpr int dir[5]={-1,0,1,0,-1};
public:
    int findPaths(int R, int C, int moves, int sr, int sc) {
        if(moves==0)
            return 0;
        int ans = (sr == 0) + (sr == R - 1) + (sc == 0) + (sc == C - 1);
        
        int dp[2][R][C];
        bool curr = false;
        fill(&dp[curr][0][0], &dp[curr][0][0] + R*C, 0);
        dp[curr][sr][sc] = 1;
        
        for(int move=1; move<moves; ++move){
            fill(&dp[!curr][0][0], &dp[!curr][0][0] + R*C, 0);
            
            for(int r=0; r<R; ++r){
                for(int c=0; c<C; ++c){
                    if(dp[curr][r][c] <= 0)
                        continue;
                    
                    for(int i=0; i<4; ++i){
                        int nr = r + dir[i],
                            nc = c + dir[i + 1];
                        if(nr < 0 || nr >= R || nc < 0 || nc >= C)
                            continue;
                        
                        dp[!curr][nr][nc] = (dp[!curr][nr][nc] + dp[curr][r][c]) % MOD;
                    }
                }
            }
            
            for(int r=0; r<R; ++r)
                ans = (0LL + ans + dp[!curr][r][0] + dp[!curr][r][C - 1]) % MOD;
            
            for(int c=0; c<C; ++c)
                ans = (0LL + ans + dp[!curr][0][c] + dp[!curr][R - 1][c]) % MOD;
            
            curr = !curr;
        }
        
        return ans;
    }
};
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