<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 dx[4]={-1,1,0,0};
    int dy[4]={0,0,1,-1};
    bool isValid(int x,int y,int n,int m){
        return (x>=0&&y>=0&&x<n&&y<m);
    }
    int countPaths(vector<vector<int>>& grid) {
        // it is just like a toposort
        //  all nodes with large values appears at last 
        // so just like dp on it
        int n=grid.size(),m=grid[0].size();
        vector<pair<int,pair<int,int>>> a;
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                a.push_back({grid[i][j],{i,j}});
            }
        }

        sort(a.begin(),a.end());

        vector<vector<int>> dp(n,vector<int>(m));
        int ans=0;
        const int M=1e9+7;
        for(int i=n*m-1;i>=0;i--){
            int x=a[i].second.first,y=a[i].second.second;
            dp[x][y]=1; // on self
            for(int k=0;k<4;k++){
                int newX=x+dx[k],newY=y+dy[k];
                if(isValid(newX,newY,n,m)&&grid[newX][newY]>grid[x][y]){
                    dp[x][y]+=dp[newX][newY];
                    dp[x][y]%=M;
                }
            }
            ans=(ans+dp[x][y])%M;
        }

        return ans;
    }
};
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