<p>You are given an <code>m x n</code> integer matrix <code>grid</code>, where you can move from a cell to any adjacent cell in all <code>4</code> directions.</p>

<p>Return <em>the number of <strong>strictly</strong> <strong>increasing</strong> paths in the grid such that you can start from <strong>any</strong> cell and end at <strong>any</strong> cell. </em>Since the answer may be very large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

<p>Two paths are considered different if they do not have exactly the same sequence of visited cells.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/05/10/griddrawio-4.png" style="width: 181px; height: 121px;" />
<pre>
<strong>Input:</strong> grid = [[1,1],[3,4]]
<strong>Output:</strong> 8
<strong>Explanation:</strong> The strictly increasing paths are:
- Paths with length 1: [1], [1], [3], [4].
- Paths with length 2: [1 -&gt; 3], [1 -&gt; 4], [3 -&gt; 4].
- Paths with length 3: [1 -&gt; 3 -&gt; 4].
The total number of paths is 4 + 3 + 1 = 8.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> grid = [[1],[2]]
<strong>Output:</strong> 3
<strong>Explanation:</strong> The strictly increasing paths are:
- Paths with length 1: [1], [2].
- Paths with length 2: [1 -&gt; 2].
The total number of paths is 2 + 1 = 3.
</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;= 1000</code></li>
	<li><code>1 &lt;= m * n &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= grid[i][j] &lt;= 10<sup>5</sup></code></li>
</ul>

const int MOD = 1e9 + 7;
const int MAXM = 1001;
const int MAXN = 1001;
int f[MAXM][MAXN];

int count(const int i, const int j, const int M, const int N, const vector<vector<int>>& grid) {
    if (f[i][j] > 0) {
        return f[i][j];
    }
    const int x = grid[i][j];
    int ret = 1;
    if (i + 1 < M && x < grid[i + 1][j]) {
        ret = (ret + count(i + 1, j, M, N, grid)) % MOD;
    }
    if (i - 1 >= 0 && x < grid[i - 1][j]) {
        ret = (ret + count(i - 1, j, M, N, grid)) % MOD;
    }
    if (j + 1 < N && x < grid[i][j + 1]) {
        ret = (ret + count(i, j + 1, M, N, grid)) % MOD;
    }
    if (j - 1 >= 0 && x < grid[i][j - 1]) {
        ret = (ret + count(i, j - 1, M, N, grid)) % MOD;
    }
    f[i][j] = ret;
    printf("[%d][%d] = %d\n", i, j, ret);
    return ret;
}

class Solution {
public:
    int countPaths(vector<vector<int>>& grid) {
        const int M = grid.size();
        if (M == 0) {
            return 0;
        }
        const int N = grid[0].size();

        memset(f, 0, sizeof(f));
        
        int ret = 0;
        for (int i = 0; i < M; i++) {
            for (int j = 0; j < N; j++) {
                ret = (ret + count(i, j, M, N, grid)) % MOD;
            }
        }
        return ret;
    }
};
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