<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:
    unordered_map<int,unordered_map<int,int>> dp;
    vector<vector<int>> a;
    int m,n;
    vector<int> dir{0,1,0,-1,0};
    int mod=1e9+7;
    int dfs(int x,int y)
    {
        if(dp.find(x)!=dp.end() && dp[x].find(y)!=dp[x].end())
         return dp[x][y];
        
        dp[x][y]=1;
       int ans=1;
        for(int i=0;i<4;i++)
        {
            int nx=x+dir[i];
            int ny=y+dir[i+1];
           
            if(min(nx,ny)>=0 && nx<m && ny<n)
            {
                if(a[nx][ny]>a[x][y])
                {
                    ans=(ans+dfs(nx,ny))%mod;
                }
            }
        }

        return dp[x][y]=ans%mod;

    }
    int countPaths(vector<vector<int>>& grid) {
        
        this->a=grid;
        this->m=a.size();
        this->n=a[0].size();
        
        int ans=0;
       

        for(int i=0;i<m;i++)
        {
            for(int j=0;j<n;j++)
            {
                ans=(ans%mod+dfs(i,j)%mod)%mod;
            };
        }

        return ans%mod;
    }
};
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