<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:
        m,n = len(grid), len(grid[0])
        mod = 10**9+7
        memo = [[-1] * n for _ in range(m)]
        cells = [(grid[i][j], i, j) for i in range(m) for j in range(n)]
        cells.sort()
        total_paths = 0

        for value, x, y in cells:
            total_paths = (total_paths + self.dfs(grid,x,y,mod,memo)) % mod
        return total_paths

    def dfs(self, grid, x, y, mod, memo):
        if memo[x][y] != -1:
            return memo[x][y]
        cnt = 1
        for dx, dy in ((-1, 0), (1, 0), (0, -1), (0, 1)):
            nx, ny = x+dx, y+dy
            if  nx < 0 or nx == len(grid) or ny < 0 or ny == len(grid[0]) or grid[nx][ny] <= grid[x][y]:
                continue
            cnt = (cnt + self.dfs(grid,nx,ny,mod,memo)) % mod
        memo[x][y] = cnt
        return memo[x][y]


1697