<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 mod = 1e9 + 7;
    vector<pair<int,int>> dir{{-1,0}, {1, 0}, {0, 1}, {0, -1}};
    int countPaths(vector<vector<int>>& arr) {
        int n = arr.size(), m = arr[0].size();
        vector<vector<long>> paths(n, vector<long>(m, 1));
        vector<vector<int>> inDegree(n, vector<int>(m, 0));
        for(int i=0; i<n; i++)
        {
            for(int j=0; j<m; j++)
            {
                for(auto [di, dj]: dir)
                {
                    int x = i + di, y = j + dj;
                    if(x>=0 && x<n && y>=0 && y<m && arr[i][j]<arr[x][y])
                        inDegree[x][y]++;
                }
            }
        }
        vector<pair<int,int>> topo;
        queue<pair<int,int>> q;
        for(int i=0; i<n; i++)
        {
            for(int j=0; j<m; j++)
                if(!inDegree[i][j]) q.push({i,j});
        }
        while(!q.empty())
        {
            auto [i,j] = q.front();
            topo.push_back({i,j});
            q.pop();
            for(auto [di, dj]: dir)
            {
                int x = i + di, y = j + dj;
                if(x>=0 && x<n && y>=0 && y<m && arr[i][j]<arr[x][y])
                {
                    if(--inDegree[x][y] == 0)
                        q.push({x, y});
                }
            }
        }
        for(auto [i,j]: topo)
        {
            for(auto [di, dj]: dir)
            {
                int x = i + di, y = j + dj;
                if(x>=0 && x<n && y>=0 && y<m && arr[i][j]<arr[x][y])
                    paths[x][y] = (paths[x][y] + paths[i][j])%mod;
            }
        }
        long ans = 0;
        for(int i=0; i<n; i++)
            for(int j=0; j<m; j++)
                ans = (ans + paths[i][j])%mod;
        return ans;
    }
};
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