<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:
    pair<int,int> dir[4] = {{0,1},{0,-1},{-1,0},{1,0}};
    int MOD = 1e9+7;
    int dfs(int r,int c, vector<vector<int>>& vis, vector<vector<int>>& dp, vector<vector<int>>& grid ){

        vis[r][c] = 1;
        int m = grid.size();
        int n = grid[0].size();

        for(int i=0; i<4; i++){
            int nr = r + dir[i].first;
            int nc = c + dir[i].second;

            if(nr>=0 && nc>=0 && nr<m && nc<n ){
                if(grid[nr][nc]>grid[r][c]){
                    if(vis[nr][nc])
                        dp[r][c] = (dp[r][c]+ dp[nr][nc])%MOD;
                    else 
                        dp[r][c] += dfs(nr,nc,vis,dp,grid);
                }
            }
        }

        return dp[r][c]%MOD;
    }
    int countPaths(vector<vector<int>>& grid) {
        int MOD = 1e9 + 7;

        int m = grid.size();
        int n = grid[0].size();

        vector<vector<int>> dp(m,vector<int>(n,1));
        // stores no. of paths if end at this index
        vector<vector<int>> vis(m,vector<int>(n,0));

        for(int i=0; i<m; i++){
            for(int j=0; j<n; j++){
                if(vis[i][j]==1) continue;

                dp[i][j] = dfs(i,j,vis,dp,grid)%MOD;

            }
        }

        long long int ans = 0;
        for(int i=0; i<m; i++){
            for(int j=0; j<n;j++){
                ans = (ans+dp[i][j]);
            }
        }

        return ans%MOD;
    }
};
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