<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:

        rows = len(grid); cols = len(grid[0])
        dp = defaultdict(int)
        def dfs(i,j):
            if dp[(i,j)] != 0:
                return dp[(i,j)]
            val1 = 0
            if i + 1 < rows and grid[i+1][j] > grid[i][j]:
                val1 = dfs(i+1,j)
            val2 = 0
            if i - 1 >= 0 and grid[i-1][j] > grid[i][j]:
                val2 = dfs(i-1,j)
            val3 = 0
            if j - 1 >= 0 and grid[i][j-1] > grid[i][j]:
                val3 = dfs(i,j-1)
            val4 = 0
            if j + 1 < cols and grid[i][j+1] > grid[i][j]:
                val4 = dfs(i,j+1)
            val = 1 + val1 + val2 + val3 + val4 
            dp[(i,j)] = val
            return val
        for i in range(rows):
            for j in range(cols):
                if dp[(i,j)] == 0:
                    dfs(i,j)
        return sum(dp.values()) % 1000000007
154026