<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>

from collections import defaultdict
class Solution:
    def countPaths(self, grid: List[List[int]]) -> int:
        total_paths = 0
        dp = defaultdict(lambda: defaultdict(lambda: 0))

        for r in range(len(grid)):
            for c in range(len(grid[0])):
                if dp[r][c] == 0:
                    total_paths += self.explore(r, c, grid, dp)
                else:
                    total_paths += dp[r][c]
        
        return total_paths % (10 ** 9 + 7)

    def explore(self, r, c, grid, dp):
        current_cell = grid[r][c]
        all_paths = 1
        for new_r, new_c in [(r+1,c), (r-1, c), (r, c-1), (r,c+1)]:
            if 0 <= new_r < len(grid) and 0 <= new_c < len(grid[0]):
                new_cell = grid[new_r][new_c]
                if new_cell > current_cell:
                    if dp[new_r][new_c] == 0:
                        all_paths += self.explore(new_r, new_c, grid, dp)
                    else:
                        all_paths += dp[new_r][new_c]
        
        dp[r][c] = all_paths
        return all_paths
            
71958