<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 dfs(vector<vector<int>>& grid, vector<vector<int>>& dp, int i, int j) {
        int mod = 1e9 + 7;
        if (dp[i][j] != -1) {
            return dp[i][j];
        }
        long long paths = 1; 
        vector<pair<int, int>> arr = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
        for (auto& dir : arr) {
            int ni = i + dir.first;
            int nj = j + dir.second;
            if (ni >= 0 && ni < grid.size() && nj >= 0 && nj < grid[0].size() 
            && grid[i][j] < grid[ni][nj]) {
                paths = (paths + dfs(grid, dp, ni, nj)) % mod;
            }
        }
        return dp[i][j] = paths;
    }
    int countPaths(vector<vector<int>>& grid) {
        ios::sync_with_stdio(false);
        cin.tie(nullptr);
        int m = grid.size();
        int n = grid[0].size();
        int mod = 1e9 + 7;
        vector<vector<int>> dp(m, vector<int>(n, -1));  
        int ans = 0;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                ans += dfs(grid, dp, i, j);
                ans %= mod;
            }
        }  
        return ans;
    }
};
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