<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>

#define MOD 1000000007
unsigned long long dp[1001][1001];

class Solution {
public:
   unsigned long long helper(vector<vector<int>>& arr, int i, int j, int prevVal) {
        // Base case: if out of bounds or value is not increasing
        if (i < 0 || j < 0 || i >= arr.size() || j >= arr[0].size() || arr[i][j] <= prevVal) 
            return 0;

        // If already computed, return the stored result
        if (dp[i][j] != -1) return dp[i][j];

        // Explore all four directions
        unsigned long long up = helper(arr, i - 1, j, arr[i][j]) % MOD;
        unsigned long long down = helper(arr, i + 1, j, arr[i][j]) % MOD;
        unsigned long long left = helper(arr, i, j - 1, arr[i][j]) % MOD;
        unsigned long long right = helper(arr, i, j + 1, arr[i][j]) % MOD;

        // Calculate total paths including the current cell
        return dp[i][j] = (1 + (up + down + left + right) % MOD) % MOD;
    }

    int countPaths(vector<vector<int>>& grid) {
        unsigned long long ans = 0;
        memset(dp, -1, sizeof(dp));

        // Iterate over each cell in the grid
        for (int i = 0; i < grid.size(); i++) {
            for (int j = 0; j < grid[0].size(); j++) {
                ans = (ans + helper(grid, i, j, -1)) % MOD;
            }
        }

        // Return the final answer with modulo
        return ans % MOD;
    }
};

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