<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:
        """
        Basically just DP.

        There is a concern here. We only mark a node as visited (in seen) *after*
        we visit it's neighboring nodes. Does this lead to an infinite loop, where
        DFS never terminates?

        No! This cannot happen, because if node y is a neighbor of node x, then
        this means grid[x] > grid[y]. So when we go to node y and then try to
        DFS from y, the node x will not be in the neighbors of y. So we will never
        be in this sort of loop.
        """
        m, n = len(grid), len(grid[0])

        res = 0
        seen = {}
        for i in range(m):
            for j in range(n):
                res += self.dfs(i, j, grid, seen)
                res = res % (10 ** 9 + 7)
        return res % (10 ** 9 + 7)

    def dfs(self, i, j, grid, seen):
        if (i, j) in seen:
            return seen[(i, j)]
    
        neighbors = self.get_neighbors(i, j, grid)
        res = 1
        for x, y in neighbors:
            res += self.dfs(x, y, grid, seen)
        seen[(i, j)] = res
        return res
        
    def get_neighbors(self, i, j, grid):
        m, n = len(grid), len(grid[0])
        curr = grid[i][j]
        res = []
        if i + 1 < m and grid[i + 1][j] < curr:
            res.append((i + 1, j))
        if i - 1 >= 0 and grid[i - 1][j] < curr:
            res.append((i - 1, j))
        if j + 1 < n and grid[i][j + 1] < curr:
            res.append((i, j + 1))
        if j - 1 >= 0 and grid[i][j - 1] < curr:
            res.append((i, j - 1))
        return res
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