<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 {
public:
    int countPaths(vector<vector<int>>& grid) {
        int n = grid.size(), m = grid[0].size(), x, y, r, c;
        int md = 1e9 + 7;
        vector<vector<int>> inDeg(n, vector<int> (m, 0));
        vector<vector<int>> store(n, vector<int> (m, 1));
        vector<pair<int, int>> dir = {{-1, 0}, {1, 0}, {0, 1}, {0, -1}};
        queue<pair<int, int>> q;

        for(int i=0;i<n;i++) {
            for(int j=0;j<m;j++) {
                for(int d = 0;d<4;d++) {
                    x = i + dir[d].first;
                    y = j + dir[d].second;
                    if(min(x, y) >= 0 && x < n && y < m && grid[x][y] > grid[i][j]) {
                        inDeg[x][y]++;
                    }
                }
            }
        }


        for(int i=0;i<n;i++) {
            for(int j=0;j<m;j++) {
                if(inDeg[i][j] == 0) {
                    q.push({i, j});
                }
            }
        }


        while(!q.empty()) {
            r = q.front().first;
            c = q.front().second;

            q.pop();

            for(int d = 0;d<4;d++) {
                x = r + dir[d].first;
                y = c + dir[d].second;

                if(min(x, y) >= 0 && x < n && y < m && grid[x][y] > grid[r][c] && inDeg[x][y] > 0) {
                    inDeg[x][y]--;
                    store[x][y] = (store[x][y] + (long long)store[r][c]) % md;
                    if(inDeg[x][y] == 0) {
                        q.push({x, y});
                    }
                }
            }
        }

        int ans = 0;

        for(int i=0;i<n;i++) {
            for(int j=0;j<m;j++) {
                ans = (ans + (long long)store[i][j]) % md;
            }
        }


        return ans;
    }
};
60214