<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>

typedef pair<int,int> pi;
class Solution {
    int rows, cols;
    vector<int> dx = {1,0,-1,0};
    vector<int> dy = {0,-1,0,1};
    const int MOD = 1e9 + 7;

    int dfs(pi val, vector<vector<int>> &d, map<pi, int> &memo) {
        memo[val] = 1;

        int depth(1);
        for(int o = 0; o < 4;o++) {
            int x = val.first + dx[o];
            int y = val.second + dy[o];
            pi next = make_pair(x,y);
            if(x < 0 || y < 0 || x >= rows || y >= cols || d[x][y] <= d[val.first][val.second])
                continue;
            // memo[val] = (memo[val] + dfs(next, d, memo)) % MOD;
            if(!memo[next]) {
                memo[val] = (memo[val] + dfs(next, d, memo)) % MOD;
            } else {
                memo[val] = (memo[val] + memo[next]) % MOD; 
            }
        }
        return memo[val];
    }
public:
    int countPaths(vector<vector<int>>& grid) {
        rows = grid.size(), cols = grid[0].size();
        map<pi, int> memo;
        int s(0);
        for(int o = 0; o < rows;o++) {
            for(int i = 0; i < cols;i++) {
                s= (s+dfs(make_pair(o,i), grid, memo)) % MOD;
            }
        }
        return s;
    }
};
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