<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 functools import lru_cache
from itertools import product
from typing import List

MOD = 10**9 + 7  # To prevent overflow


class Solution:
    def countPaths(self, grid: List[List[int]]) -> int:
        self.grid = grid
        self.m = len(grid)
        self.n = len(grid[0])

        return (
            sum(self.visit(x, y) for x, y in product(range(self.m), range(self.n)))
            % MOD
        )

    @lru_cache(None)
    def visit(self, x: int, y: int) -> int:
        neighbours = [
            (x - 1, y) if x > 0 else None,
            (x + 1, y) if x < self.m - 1 else None,
            (x, y - 1) if y > 0 else None,
            (x, y + 1) if y < self.n - 1 else None,
        ]
        valid_paths = [
            (nx, ny)
            for n in neighbours
            if n is not None
            and (nx := n[0], ny := n[1])
            and self.grid[nx][ny] > self.grid[x][y]
        ]

        return (1 + sum(self.visit(nx, ny) for nx, ny in valid_paths)) % MOD

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