<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>

int mod = 1000000007;
class Solution {
public:
    int paths(int i, int j, vector<vector<int>>& grid, vector<vector<int>>& dp) {
        int n = grid.size();
        int m = grid[0].size();
        if(i < 0 || i >= n || j < 0 || j >= m) return 0;
        if(dp[i][j] != -1) return dp[i][j];
        int answer = 1;
        vector<int> di{1, -1, 0, 0};
        vector<int> dj{0, 0, 1, -1};
        for(int k = 0; k < 4; k++) {
            int iNew = i + di[k]; 
            int jNew = j + dj[k];
            if(iNew < 0 || iNew >= n || jNew < 0 || jNew >= m) continue;
            if(grid[iNew][jNew] <= grid[i][j]) continue;
            answer = (answer + paths(iNew, jNew, grid, dp))%mod;
        }
        return (dp[i][j] = answer);
    }
    int countPaths(vector<vector<int>>& grid) {
        int n = grid.size();
        int m = grid[0].size();
        vector<vector<int>> dp(n, vector<int>(m, -1));
        int answer = 0;
        for(int i = 0; i < n; i++)
            for(int j = 0; j < m; j++)
                answer = (answer + paths(i, j, grid, dp))%mod;
        return answer;
    }
};
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