<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:
        # из исходной матрицы нужно получить номера каждой ячейки: grid[i][j] => adj[a]
        # (иначе придется обрасти словарями);
        # выявить какой номер сосед какому;
        # в соответствии с условием(значение соседа дб больше значения узла) собрать список смежности;
        # в списке путей вести учет количества путей для каждого узла;

        m = len(grid)
        n = len(grid[0])
        # Since the answer may be very large, return it modulo 1e9 + 7:
        mod = int(1e9 + 7)

        # create adjacency list from matrix:
        adj = [[] for _ in range(m*n)]
        neighbours = ((-1, 0), (0, 1), (1, 0),(0, -1))
        for i in range(m):
            for j in range(n):
                for x, y in neighbours:
                    a, b = i+x, j+y
                    if a in range(0, m) and b in range(0, n) and grid[i][j] < grid[a][b]:
                        adj[n*i+j].append(n*a+b)

        # create indegree list:
        l = len(adj)
        indegree = [0]*l
        for i in range(l):
            for v in adj[i]:
                indegree[v] += 1

        paths = [0]*m*n  # number of paths from each vertex
        deq = collections.deque()
        for i in range(l):
            if indegree[i] == 0:
                deq.append(i)

        while deq:
            v = deq.popleft()
            paths[v] += 1

            for u in adj[v]:
                paths[u] += paths[v]
                paths[u] %= mod
                indegree[u] -= 1
                if indegree[u] == 0:
                    deq.append(u)

        return sum(paths) % mod
2131