<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 m,n;
    int mod=1e9+7;
    vector<vector<int>>directions{{-1,0},{1,0},{0,-1},{0,1}};
    int recurssion(int row,int col,vector<vector<int>>& matrix,vector<vector<int>>&dp){
        if(dp[row][col]!=-1) return dp[row][col];
        int count=1;
        for(auto it:directions){
            int x=row+it[0];
            int y=col+it[1];
            if(x>=0 && y>=0 && x<m && y<n && matrix[x][y]>matrix[row][col]){
                count=(count+recurssion(x,y,matrix,dp))%mod;
            }
        }
        return dp[row][col]=count%mod;

        
    }
    int countPaths(vector<vector<int>>& matrix) {
        m=matrix.size();
        n=matrix[0].size();
        int count=0;
        vector<vector<int>>dp(m,vector<int>(n,-1));
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                count=(count+recurssion(i,j,matrix,dp))%mod;
                
            }
        }
        
        return count%mod;
        
    }
};
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