<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>

class Solution {
public:
    int n;
    int m;
    int t[1001][1001];
    vector<vector<int>>directions{{0,1}, {0,-1}, {-1,0}, {1,0}};
    long long M = 1e9 + 7;
    bool issafe(int i, int j)
    {
        return (i >= 0 && i < m && j >= 0 && j < n);
    }
    int solve(int i, int j, vector<vector<int>>& grid)
    {
        if(t[i][j] != -1)
            return t[i][j];
        int ans = 1;
        for(vector<int>&dir: directions)
        {
            int i_ = i + dir[0];
            int j_ = j + dir[1];
            if(issafe(i_, j_) && grid[i][j] < grid[i_][j_])
                ans = (ans + solve(i_, j_, grid)) % M;
        }
        return  t[i][j] = ans;
    }
    int countPaths(vector<vector<int>>& grid) {
        m = grid.size();
        n = grid[0].size();
        int result = 0;
        memset(t, -1, sizeof(t));
        for(int i = 0; i < m; i++)
        {
            for(int j = 0; j < n; j++)
            {
                result = (result + solve(i, j, grid)) % M;
            }
        }
        return result;
    }
};
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