<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:
    #define ll long long
    #define pll pair<ll, ll>
    const ll MOD = 1e9 + 7;
    vector<vector<pll>> dp;
    int n, m;
    pll mx = {LLONG_MIN, LLONG_MIN};
    pll mn = {LLONG_MAX, LLONG_MAX};

    pll solve(int i, int j, vector<vector<int>>& grid) {
        if (i >= n || j >= m)
            return mx;
        if (i == n - 1 && j == m - 1)
            return {grid[i][j], grid[i][j]};
        if (dp[i][j] != mx)
            return dp[i][j];
        pll down = solve(i + 1, j, grid);
        pll right = solve(i, j + 1, grid);
        ll maxi = LLONG_MIN, mini = LLONG_MAX;
        
        if (down != mx) {
            maxi = max(grid[i][j] * down.first, grid[i][j] * down.second);
            mini = min(grid[i][j] * down.first, grid[i][j] * down.second);
        }
        if (right != mx) {
            maxi = max({maxi, grid[i][j] * right.first, grid[i][j] * right.second});
            mini = min({mini, grid[i][j] * right.first, grid[i][j] * right.second});
        }
        
        return dp[i][j] = {mini, maxi};
    }

    int maxProductPath(vector<vector<int>>& grid) {
        n = grid.size();
        m = grid[0].size();
        if (n == 1 && m == 1)
            return grid[0][0] % MOD;
        dp.resize(n, vector<pll>(m, mx));
        ll ans = solve(0, 0, grid).second;
        return ans < 0 ? -1 : ans % MOD;
    }
};

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