<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 n = grid.size();
        int m = grid[0].size();
        vector<vector<long long>> dp(n, vector<long long>(m, 1));
        priority_queue<pair<int, pair<int, int>>, vector<pair<int, pair<int, int>>>, greater<pair<int, pair<int, int>>>> pq;
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                pq.push({grid[i][j],{i, j}});
            }
        }
        int M = 1e9 + 7;
        while(!pq.empty()){
            int i = pq.top().second.first;
            int j = pq.top().second.second;
            int num = pq.top().first;
            pq.pop();
            vector<int> ab = {-1, 0, 1, 0};
            vector<int> bc = {0, -1, 0, 1};
            for(int k=0;k<4;k++){
                int r = i+ab[k];
                int c = j+bc[k];
                if(r < n && r >= 0 && c >= 0 && c < m){
                    if(grid[i][j] < grid[r][c]){
                        dp[r][c] = (dp[r][c]%M + dp[i][j]%M)%M;
                    }
                }
            }
        }
        // for(int i=0;i<n;i++){
        //     for(int j=0;j<m;j++){
                // vector<int> ab = {-1, 0, 1, 0};
                // vector<int> bc = {0, -1, 0, 1};
                // for(int k=0;k<4;k++){
                //     int r = i+ab[k];
                //     int c = j+bc[k];
                //     if(r < n && r >= 0 && c >= 0 && c < m){
                        
                //     }
                // }
        //     }
        // }
        long long ans = 0;
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                ans = (ans%M+dp[i][j]%M)%M;
                // cout<<dp[i][j]<<" ";
            }
            // cout<<endl;
        }
        return ans%M;
    }
};
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