<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 MOD = 1e9 + 7;
    int countPaths(vector<vector<int>>& grid) {
        int m = grid.size(), n = grid[0].size();
        vector<vector<int>> dp(m, vector<int>(n, 1));
        vector<pair<int, int>> cells;
        
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                cells.push_back({i, j});
            }
        }
        
        sort(cells.begin(), cells.end(), [&](pair<int, int>& a, pair<int, int>& b) {
            return grid[a.first][a.second] < grid[b.first][b.second];
        });
        
        int result = 0;
        int dirs[5] = {0, 1, 0, -1, 0};
        
        for (auto& cell : cells) {
            int i = cell.first, j = cell.second;
            for (int d = 0; d < 4; ++d) {
                int ni = i + dirs[d], nj = j + dirs[d + 1];
                if (ni >= 0 && ni < m && nj >= 0 && nj < n && grid[ni][nj] > grid[i][j]) {
                    dp[ni][nj] = (dp[ni][nj] + dp[i][j]) % MOD;
                }
            }
            result = (result + dp[i][j]) % MOD;
        }
        
        return result;
    }
};
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