<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(),m = grid[0].size();
        vector<vector<int>> dp (n,vector<int>(m,1));
        map<int,vector<pair<int,int>>> M;
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                M[grid[i][j]].emplace_back(i,j);
            }
        } 
        int ans=0;
        for(auto it = M.rbegin();it != M.rend();it++){
            for(auto item2 : it->second){
                int r = item2.first;
                int c = item2.second;
                if(r+1<n &&grid[r][c]<grid[r+1][c])  dp[r][c] += dp[r+1][c];
                if(r-1>=0 &&grid[r][c]<grid[r-1][c]) dp[r][c] += dp[r-1][c];
                if(c+1<m &&grid[r][c]<grid[r][c+1])  dp[r][c] += dp[r][c+1];
                if(c-1>=0 &&grid[r][c]<grid[r][c-1]) dp[r][c] += dp[r][c-1];
                dp[r][c]%=1000000007;
                ans+=dp[r][c];
                ans%=1000000007;
            }
        }
        return ans;
    }
};
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