<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>

using ll = long long;
class Solution {
public:
    int countPaths(vector<vector<int>>& grid) {
        ll n = grid.size(), m = grid[0].size(), ans = 0, mod = 1e9+7;
        vector<vector<ll>> dp(n + 1, vector<ll> (m + 1, -1));
        auto add = [&](ll a, ll b){
            return ((a%mod) + (b%mod))%mod;
        };
        function<ll(int i, int j)> dfs;
        dfs = [&](int i, int j){
            if(dp[i][j] != -1)return dp[i][j];
            ll res = 0;
            if(i - 1 >= 0 && grid[i - 1][j] > grid[i][j])res = add(res, dfs(i - 1, j));
            if(j - 1 >= 0 && grid[i][j - 1] > grid[i][j])res = add(res, dfs(i, j - 1));
            if(j + 1 < m && grid[i][j + 1] > grid[i][j])res = add(res, dfs(i, j + 1));
            if(i + 1 < n && grid[i + 1][j] > grid[i][j])res = add(res, dfs(i + 1, j));
            return dp[i][j] = res + 1;
        };
        for(int i = 0; i < n; i++){
            for(int j = 0; j < m; j++){
                if(dp[i][j] == -1)dfs(i, j);
                ans = add(ans, dp[i][j]);
            }
        }
        return ans;
    }
};
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