<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 {
    vector<vector<int>> dp;
    vector<vector<int>> moves = {{0,1},{0,-1},{1,0},{-1,0}};
    int mod = 1e9+7;
public:
    int dfs(vector<vector<int>> &grid,int i,int j,vector<vector<bool>> &visited) {
        int m = grid.size(), n = grid[0].size();
        if(dp[i][j]!=-1) return dp[i][j];
        visited[i][j] = true;
        int res = 1;
        for(auto &x:moves) {
            int in = i + x[0], jn = j + x[1];
            if(in>=0 and jn>=0 and in<m and jn<n and !visited[in][jn] and grid[in][jn]>grid[i][j]) {
                res = (res + dfs(grid,in,jn,visited))%mod;
            }
        }
        visited[i][j] = false;
        return dp[i][j] = res%mod;
    }

    int countPaths(vector<vector<int>>& grid) {
        int m = grid.size(), n = grid[0].size();
        dp = vector<vector<int>> (m,vector<int> (n,-1));
        vector<vector<bool>> visited(m,vector<bool> (n,false));
        for(int i = 0; i<m; i++) {
            for(int j = 0; j<n; j++) {
                if(dp[i][j]==-1) {
                    dfs(grid,i,j,visited);
                }
            }
        }
        int res = 0;
        for(auto &x:dp) {
            for(auto &y:x) {
                res = (res+y)%mod;
            }
        }
        return res%mod;
    }
};
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