<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>

int dir[][2] = {{0,1},{1,0},{-1,0}, {0,-1}};

int mod = 1e9 +7;
vector<vector<int>> dp;
int n, m;
class Solution {

public:
    int solve(int i, int j, vector<vector<int>>& matrix) {
      if(dp[i][j] != -1) return dp[i][j];
      int &ans = dp[i][j];
       ans = 0;
       for(int d = 0 ; d < 4; ++d) { 
            int ni  = i + dir[d][0], nj = j + dir[d][1]; 
            if(ni < m && ni >=0 && nj >=0 && nj < n && matrix[i][j] < matrix[ni][nj]) {
              ans = (ans + (1 + solve(ni, nj, matrix))) % mod;
            }
       }
       return ans; 
    }
    int countPaths(vector<vector<int>>& matrix) {
        m = matrix.size(), n = matrix[0].size();
       dp.assign(m , vector<int>(n, -1));
      
       int ans = (n * m) % mod;
       for(int i = 0; i < m;++i) {
         for(int j = 0; j < n;++j) {
            ans = (ans + solve(i,j, matrix))% mod;
           // cout << i << " " << j << " "<< ans << "\n";
         }
       }
 
        return ans;
    }
};
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