<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:
    #define MOD (int)(1E9 + 7)
    vector<int> axis{ 0, -1, 0, 1, 0 };
    int m, n;
    int dp[1001][1001];
    int dfs(int i, int j, vector<vector<int>>& matrix, int prev) {
        if(i < 0 || j < 0 || i == m || j == n || matrix[i][j] <= prev) return 0;
        if(dp[i][j] != -1) return dp[i][j];
        int ans = 0;
        for(int idx = 1; idx < axis.size(); idx++) {
            ans = (ans + dfs(i + axis[idx - 1], j + axis[idx], matrix, matrix[i][j])) % MOD;
        }
        return dp[i][j] = ans + 1;
    }
    int countPaths(vector<vector<int>>& matrix) {
        m = matrix.size(), n = matrix[0].size();
        memset(dp, -1, sizeof dp);
        int ans = 0;
        for(int i = 0; i < m; i++) {
            for(int j = 0; j < n; j++) {
                ans = (ans + dfs(i, j, matrix, -1)) % MOD;
            }
        }
        return ans;
    }
};
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