<p>You are given a <code>m x n</code> matrix <code>grid</code>. Initially, you are located at the top-left corner <code>(0, 0)</code>, and in each step, you can only <strong>move right or down</strong> in the matrix.</p>

<p>Among all possible paths starting from the top-left corner <code>(0, 0)</code> and ending in the bottom-right corner <code>(m - 1, n - 1)</code>, find the path with the <strong>maximum non-negative product</strong>. The product of a path is the product of all integers in the grid cells visited along the path.</p>

<p>Return the <em>maximum non-negative product <strong>modulo</strong> </em><code>10<sup>9</sup> + 7</code>. <em>If the maximum product is <strong>negative</strong>, return </em><code>-1</code>.</p>

<p>Notice that the modulo is performed after getting the maximum product.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/12/23/product1.jpg" style="width: 244px; height: 245px;" />
<pre>
<strong>Input:</strong> grid = [[-1,-2,-3],[-2,-3,-3],[-3,-3,-2]]
<strong>Output:</strong> -1
<strong>Explanation:</strong> It is not possible to get non-negative product in the path from (0, 0) to (2, 2), so return -1.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/12/23/product2.jpg" style="width: 244px; height: 245px;" />
<pre>
<strong>Input:</strong> grid = [[1,-2,1],[1,-2,1],[3,-4,1]]
<strong>Output:</strong> 8
<strong>Explanation:</strong> Maximum non-negative product is shown (1 * 1 * -2 * -4 * 1 = 8).
</pre>

<p><strong class="example">Example 3:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/12/23/product3.jpg" style="width: 164px; height: 165px;" />
<pre>
<strong>Input:</strong> grid = [[1,3],[0,-4]]
<strong>Output:</strong> 0
<strong>Explanation:</strong> Maximum non-negative product is shown (1 * 0 * -4 = 0).
</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;= 15</code></li>
	<li><code>-4 &lt;= grid[i][j] &lt;= 4</code></li>
</ul>

class Solution {
public:

    pair<long long, long long> recur(vector<vector<int>>&grid, vector<vector<pair<long long,long long>>>&dp, int i, int j){
        int n=grid.size();
        int m=grid[0].size();
        if(i==n-1 && j==m-1)return {grid[i][j], grid[i][j]};
        if(dp[i][j].first != LLONG_MIN)return dp[i][j];
        long long mx = LLONG_MIN, mn = LLONG_MAX;
        int mod = (int)1e9+7;
        if(i+1<n){
            pair<long long, long long>temp = recur(grid, dp, i+1, j);
            long long val1 = (temp.first*grid[i][j]);
            long long val2 = (temp.second*grid[i][j]);
            mn = min({mn, val1, val2});
            mx = max({mx, val1, val2});
        }
        if(j+1<m){
            pair<long long, long long>temp = recur(grid, dp, i, j+1);
            long long val1 = (temp.first*grid[i][j]);
            long long val2 = (temp.second*grid[i][j]);
            mn = min({mn, val1, val2});
            mx = max({mx, val1, val2});
        }

        return dp[i][j]={mx, mn};
    }
    int maxProductPath(vector<vector<int>>& grid) {
        int n=grid.size();
        int m=grid[0].size();
        int mod = (int)1e9+7;
        vector<vector<pair<long long,long long>>>dp(n,vector<pair<long long,long long>>(m,{LLONG_MIN, LLONG_MIN}));
        pair<long long, long long>ans = recur(grid, dp, 0, 0);
        return max(ans.first, ans.second)<0 ? -1:max(ans.first, ans.second)%mod;
    }
};
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