<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:
    void dfs(int i,int j,vector<vector<int>>&vis,vector<vector<vector<pair<int,int>>>>&adj,stack<pair<int,int>>&st)
    {
        //cout<<i<<" "<<j<<endl;
        vis[i][j]=1;
        for(auto it:adj[i][j])
        {
            int i1=it.first,j1=it.second;
            if(!vis[i1][j1])
            dfs(i1,j1,vis,adj,st);
        }
        st.push({i,j});
    }
    int countPaths(vector<vector<int>>& grid) {
        int e=1e9+7;
        int m=grid.size(),n=grid[0].size();
        vector<vector<vector<pair<int,int>>>>adj(m,vector<vector<pair<int,int>>>(n));
        vector<vector<int>>vis(m,vector<int>(n,0));
       // cout<<"gi";
        for(int i=0;i<m;i++)
        {
            for(int j=0;j<n;j++)
            {   
                if(i-1>=0&&grid[i-1][j]>grid[i][j]) 
                adj[i][j].push_back({i-1,j});
                if(i+1<m&&grid[i+1][j]>grid[i][j])
                adj[i][j].push_back({i+1,j});
                if(j-1>=0&&grid[i][j-1]>grid[i][j])
                adj[i][j].push_back({i,j-1});
                if(j+1<n&&grid[i][j+1]>grid[i][j])
                adj[i][j].push_back({i,j+1});
            }
        }
       // cout<<"hi";
        stack<pair<int,int>>st;
        vector<pair<int,int>>topo;
        vector<vector<int>>dp(m,vector<int>(n,1));
        for(int i=0;i<m;i++)
        {
            for(int j=0;j<n;j++)
            {
                if(vis[i][j]==1) continue;
                dfs(i,j,vis,adj,st);
            }
        }
      //  cout<<"ji";
        while(!st.empty())
        {
            topo.push_back(st.top());
            st.pop();
        }
        int ans=0;
        for(auto it:topo)
        {
            for(auto n:adj[it.first][it.second])
            {
                dp[n.first][n.second]=(dp[n.first][n.second]+(dp[it.first][it.second]))%e;
            }
            ans=(ans+dp[it.first][it.second])%e;
        }
        return (ans)%e;
    }
};
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