<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>

ans = None
class Solution:
    def countPaths(self, grid: List[List[int]]) -> int:
        global ans
        m = len(grid)
        n = len(grid[0])
        memory = {}
        ans = 0
        def dfs(i,j):
            global ans
            val = grid[i][j]
            if (i,j) in memory:
                return memory[(i,j)]
            # print(val)
            up = i-1 if i>0 else None
            down = i+1 if i<m-1 else None
            left = j-1 if j>0 else None
            right = j+1 if j<n-1 else None
            # print(val,up,down,left,right)
            a = b = c = d = 0
            if left!=None and grid[i][left]>val:
                # print(val,"a")
                a = dfs(i,left)
            if right!=None and grid[i][right]>val:
                # print(val,"b")
                b = dfs(i,right)
            if up!=None and grid[up][j]>val:
                # print(val,"c")
                c = dfs(up,j)
            if down!=None and grid[down][j]>val:
                # print(val,"d")
                d = dfs(down,j)

            ret_val = a+b+c+d+1
            memory[(i,j)] = ret_val
            ans+= ret_val
            # print(val,":",a,b,c,d,":",ret_val)
            return ret_val
            # print(up,down,left,right)

        for i in range(m):
            for j in range(n):
                dfs(i,j)
        # print(memory)
        return ans % (10 ** 9 + 7)
1562