<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 {
const long long M=1e9+7;
vector<long long>d1 ={-1,1,0,0};
vector<long long>d2 ={0,0,1,-1};
vector<vector<long long>>visited;
vector<vector<long long>>dp;
long long ans;
void dfs(long long i, long long j, vector<vector<vector<pair<long long,long long>>>>&adj){
    visited[i][j]=1;
    for(auto p: adj[i][j]){
        if(!visited[p.first][p.second]){
            dfs(p.first,p.second,adj);
        }
        dp[i][j]+=dp[p.first][p.second];
        dp[i][j]%=M;
    }
    ans+=dp[i][j];
    ans%=M;
}
public:
    int countPaths(vector<vector<int>>& grid) {
        ans=0;
        long long n=grid.size();
        long long m=grid[0].size();
        vector<vector<vector<pair<long long,long long>>>>adj(n,vector<vector<pair<long long,long long>>>(m));
        for(long long i=0;i<n;i++){
            for(long long j=0;j<m;j++){
                for(long long k=0;k<4;k++){
                    long long i1=i+d1[k];
                    long long j1=j+d2[k];
                    if(i1>=0 && i1<n && j1>=0 && j1<m && grid[i1][j1]<grid[i][j]){
                        adj[i][j].push_back({i1,j1});
                    }
                }
            }
        }
        visited.resize(n,vector<long long>(m,0));
        dp.resize(n,vector<long long>(m,1));
        for(long long i=0;i<n;i++){
            for(long long j=0;j<m;j++){
                if(!visited[i][j]){
                    dfs(i,j,adj);
                }
            }
        }
        ans=ans%M;
        return (int)ans;        
    }
};
155026