<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:
        Mod = 10**9+7  # Define the modulus
        n, m = len(grid), len(grid[0])  # Get the size of the grid
        row = [-1 for _ in range(m)]  # Initialize the visited state of each row as -1
        visited = [row[:] for _ in range(n)]  # Initialize the visited state of the entire grid as -1

        def dfs(i, j):
            if visited[i][j] != -1:  # If the position has been visited before, directly return the visited result
                return visited[i][j]
            result = 1  # Initial result is 1, representing the number of paths from the current position to the destination
            adj = [(i + 1, j), (i, j + 1), (i - 1, j), (i, j - 1)]  # Define the four directions for adjacent positions
            for a, b in adj:
                if 0 <= a < n and 0 <= b < m:  # Make sure the adjacent position is within the grid
                    if grid[a][b] > grid[i][j]:  # If the height of the adjacent position is greater than the current position, continue exploring in that direction
                        result = (result + dfs(a, b)) % Mod  # Update the result and take the modulus
            visited[i][j] = result  # Save the result to the visited state to avoid duplicate calculations
            return result

        result = 0  # Initial result is 0
        for i in range(n):
            for j in range(m):
                if visited[i][j] == -1:  # If the position has not been visited yet, perform depth-first search
                    visited[i][j] = dfs(i, j)  # Update the visited result of that position
                result = (result + visited[i][j]) % Mod  # Add the visited result of that position to the final result and take the modulus

        return result  # Return the final result
31835