<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 = 1e9+7;
    bool inRange(vector<vector<int>> &grid, int i , int j){
        int n = grid.size();
        int m = grid[0].size();
        if(i>=0 && j >=0 && i <n && j <m){
            return true;
        }
        return false;
    }
    bool checkMax(vector<vector<int>> &grid, int x , int y){
        
        vector <int> dx = {1,-1,0,0};
        vector <int> dy = {0,0,1,-1};
        for(int i = 0 ;i < 4 ; i++){
            int xdash = x + dx[i];
            int ydash = y + dy[i];
            if(inRange(grid,xdash,ydash)){
                if(grid[xdash][ydash]>grid[x][y]){
                    return false;
                }
            }
        }
        return true;
    }
    int dfs(vector<vector<int>> &grid, int x , int y, vector<vector<int>> &dp){
        int i = x;
        int j = y;
        if(dp[i][j]!=-1){
            return dp[i][j];
        }
        if(checkMax(grid,i,j)){
            dp[i][j] = 1;
            return 1;
        }
        vector <int> dx = {1,-1,0,0};
        vector <int> dy = {0,0,1,-1};
        int total = 1;
        for(int i = 0 ;i < 4 ; i++){
            int xdash = x + dx[i];
            int ydash = y + dy[i];
            if(inRange(grid,xdash,ydash) && grid[xdash][ydash]>grid[x][y]){
                total = (total%M +dfs(grid,xdash,ydash,dp)%M)%M;
            }
        }
        dp[i][j] = total;
        return total;
    }
    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 ans = 0 ;
        for(int i = 0 ; i < n ; i++){
            for(int j = 0 ; j < m ; j++){
                ans= (ans%M + dfs(grid,i,j,dp)%M)%M;
            }
        }
        return ans;
    }
};
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