<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:
        mat = []
        for i in range(0,len(grid)):
            mat.append([])
            for j in range(0, len(grid[0])):
                mat[i].append(1)
        visited=set()
        total = 0
        for i in range(0,len(grid)):
            for j in range(0,len(grid[0])):
                total += self.traverse(i,j,grid,mat,visited)
        #print(mat)
        return total % (pow(10,9)+7)

    def traverse(self, i,j,grid,mat,visited):
        if (i,j) in visited:
            return mat[i][j]
        cells = self.adj_cells(i,j,len(grid),len(grid[0]))
        for each_i, each_j in cells:
            if grid[i][j] < grid[each_i][each_j]:
                if not mat[each_i][each_j] == 0:
                    mat[i][j] += self.traverse(each_i,each_j,grid, mat, visited)
                else:
                    mat[i][j] += mat[each_i][each_j]
        visited.add((i,j))
        return mat[i][j]
    
    def adj_cells(self, i,j, m,n):
        cells = []
        if i+1<m:
            cells.append((i+1,j))
        if j+1<n:
            cells.append((i,j+1))
        if i-1>=0:
            cells.append((i-1,j))
        if j-1>=0:
            cells.append((i,j-1))
        return cells

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