<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 countPaths(vector<vector<int>>& grid) {
        int m = grid.size(), n = grid[0].size();

        int DIR[5] = {0,1,0,-1,0}; 
        int mod = 1e9 + 7; 
        vector<vector<int>> adj(n*m);
        vector<int> indeg(n*m, 0), dp(n*m, 1);

        for(int i=0; i<m; i++){
            for(int j=0; j<n; j++){
                int ind = i * n + j; 
                for(int k=0; k<4; k++){
                    int nr = i + DIR[k]; 
                    int nc = j + DIR[k+1]; 
                    if(nr>=0 && nc>=0 && nr<m && nc<n && grid[i][j] < grid[nr][nc]){
                        int new_ind = nr*n + nc; 
                        adj[ind].push_back(new_ind);  
                        indeg[new_ind]++; 
                    }
                }
            }
        }

        int ans = 0;
        queue<int> q;
        for(int i=0; i<n*m; i++){
            if(indeg[i] == 0) q.push(i);
        }
        while(q.size()){
            int node = q.front();
            q.pop();
            ans = (ans + dp[node])%mod;

            for(auto it: adj[node]){
                dp[it] = (dp[it] +  dp[node]) % mod;
                indeg[it]--;
                if(indeg[it] == 0) q.push(it);
            }        
        }
        return ans;
    }
};
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