<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 __init__(self):
        self.cost = []
        self.ans = 0
        
    
    def countPaths(self, grid: list[list[int]]) -> int:
        n, m = len(grid), len(grid[0])
        self.cost = [[-1]*m for i in range(n)]
        vert = []
        for i in range(n):
            for j in range(m):
                vert.append((grid[i][j], i, j))
        srdVert = sorted(vert, key=lambda x: x[0])
        p = len(srdVert)
        for k in range(p-1, -1, -1):
            val, i, j = srdVert[k]
            pairs = [(min(i+1, n-1), j), (max(i-1, 0), j), (i, min(j+1, m-1)), (i, max(j-1, 0))]
            self.cost[i][j] = 1
            ln = 0
            for x, y in pairs:
                if (x, y) != (i, j) and self.cost[x][y] > -1 and grid[x][y] > grid[i][j]:
                    ln += self.cost[x][y]
            self.cost[i][j] += ln
            self.ans += self.cost[i][j]
        return self.ans % (10**9 + 7)
        
73781