<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>

typedef long long int lli;
typedef pair<int,int> pii;
typedef pair<long long int,long long int> plii;

class Solution {
public:
    int maxProductPath(vector<vector<int>>& grid) {
        int m= grid.size(), n= grid[0].size();
        int mod= 1e9+7;
        vector<vector<plii>> val(m, vector<plii>(n));

        for(int i=m-1;i>=0;i--){
            for(int j=n-1; j>=0; j--){
                lli x= grid[i][j];
                vector<lli> values;
                if(j+1<n) {
                    values.push_back(x*val[i][j+1].first);
                    values.push_back(x*val[i][j+1].second);
                }
                if(i+1 <m){
                    values.push_back(x*val[i+1][j].first);
                    values.push_back(x*val[i+1][j].second);
                }
                int nValues= values.size();
                if(nValues>0){
                    //sort maximum 4 values..after sort, values[0]= most -ve and values[values.size()-1]= most +ve
                    sort(values.begin(), values.end());
                    val[i][j]= {values[0], values[nValues-1]};
                }
                else{
                    val[i][j]= {x, x}; //this occurs only for bottom-right corner cell
                }
            }
        }
        if(val[0][0].second <0) return -1; //if most +ve is < 0, then return -1
        return (val[0][0].second) %mod; //return most +ve
    }
};
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