<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:
        self.record = [[-1 for _ in grid[0]] for r in grid]
        self.dirs = [(0,1),(0,-1),(1,0),(-1,0)]
        self.h, self.w = len(grid), len(grid[0])
        total = 0
        for i in range(self.h):
            for j in range(self.w):
                total += self.dfs(i,j, grid)
                if total > 1000000000:
                    total = total % (1000000000+7)
        return total

    def dfs(self, i,j,matrix):
        # return num of possible paths starting from i,j
        if self.record[i][j]>0:
            return self.record[i][j]
        own = 1
        from_neigh = 0
        for d in self.dirs:
            new_i, new_j = i+d[0], j+d[1]
            if new_i<0 or new_i>=self.h or new_j<0 or new_j>=self.w:
                continue
            if matrix[new_i][new_j] <= matrix[i][j]:
                continue
            from_neigh += self.dfs(new_i, new_j, matrix)
        l = own+from_neigh
        # if l > 10:
        #     l = l % 10
        self.record[i][j] = l
        return l
68310