<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:
    def countPaths(self, grid: List[List[int]]) -> int:
        mod = 10 ** 9 + 7
        adj_list = {(i,j): set() for i in range(len(grid)) for j in range(len(grid[0]))}
        for i in range(len(grid)):
            if i > 0 and grid[i-1][0] > grid[i][0]:
                adj_list[(i,0)].add((i-1,0))
            if i < len(grid)-1 and grid[i+1][0] > grid[i][0]:
                adj_list[(i,0)].add((i+1,0))
            if len(grid[0]) > 1 and grid[i][1] > grid[i][0]:
                adj_list[(i,0)].add((i,1))
        for j in range(len(grid[0])):
            if j > 0 and grid[0][j-1] > grid[0][j]:
                adj_list[(0,j)].add((0, j-1))
            if j < len(grid[0])-1 and grid[0][j+1] > grid[0][j]:
                adj_list[(0,j)].add((0, j+1))
            if len(grid) > 1 and grid[1][j] > grid[0][j]:
                adj_list[(0,j)].add((1,j))
        for i in range(1, len(grid)):
            for j in range(1, len(grid[0])):
                if grid[i-1][j] > grid[i][j]:
                    adj_list[(i,j)].add((i-1,j))
                if i < len(grid)-1 and grid[i+1][j] > grid[i][j]:
                    adj_list[(i,j)].add((i+1,j))
                if grid[i][j-1] > grid[i][j]:
                    adj_list[(i,j)].add((i,j-1))
                if j < len(grid[0])-1 and grid[i][j+1] > grid[i][j]:
                    adj_list[(i,j)].add((i,j+1))
        top_sorted = []
        visited = set()
        def dfs(curr):
            if curr not in visited:
                visited.add(curr)
                for c in adj_list[curr]:
                    dfs(c)
                top_sorted.append(curr)
        for i in range(len(grid)):
            for j in range(len(grid[0])):
                if (i,j) not in visited:
                    dfs((i,j))
        top_sorted.reverse()
        #dp[(i,j)] = number of strictly increasing paths ending at (i,j)
        dp = {(i,j) : 1 for i in range(len(grid)) for j in range(len(grid[0]))}
        for start in top_sorted:
            for c in adj_list[start]:
                dp[c] += dp[start]    
        return sum([dp[k] % mod for k in dp]) % mod

        
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