<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();
        vector<vector<long long>> dp(n, vector<long long>(m, 0));
        
        auto customOp = [&grid] (const pair<int, int> l, const pair<int, int> r) -> bool {
            return grid[l.first][l.second] > grid[r.first][r.second];
        };
        priority_queue<pair<int, int>, vector<pair<int, int>>, decltype(customOp)> pq(customOp);
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                pq.emplace(i, j);
            }
        }
        long long res = 0;
        while (!pq.empty()) {
            auto v = pq.top();
            pq.pop();
            int x = v.first, y = v.second;
            if (x+1<n && grid[x][y] > grid[x+1][y]) {
                dp[x][y] += dp[x+1][y]+1;
                dp[x][y] = dp[x][y] % (int)(pow(10, 9) + 7);
            }
            if (x-1>=0 && grid[x][y] > grid[x-1][y]) {
                dp[x][y] += dp[x-1][y]+1;
                dp[x][y] = dp[x][y] % (int)(pow(10, 9) + 7);
            }
            if (y+1<m && grid[x][y] > grid[x][y+1]) {
                dp[x][y] += dp[x][y+1]+1;
                dp[x][y] = dp[x][y] % (int)(pow(10, 9) + 7);
            }
            if (y-1>=0 && grid[x][y] > grid[x][y-1]) {
                dp[x][y] += dp[x][y-1]+1;
                dp[x][y] = dp[x][y] % (int)(pow(10, 9) + 7);
            }
            res += dp[x][y];
            res = res % (int)(pow(10, 9) + 7);
        }
        res += (n * m);
        return res % (int)(pow(10, 9) + 7);

    }
};
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