<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:
    def maxProductPath(self, grid: List[List[int]]) -> int:
        m, n = len(grid), len(grid[0])

        @lru_cache(None)
        def fn(i, j):
            if i == 0 and j == 0: return grid[0][0], grid[0][0]
            if i < 0 or j < 0: return -inf, inf
            if grid[i][j] == 0: return 0, 0
            mx1, mn1 = fn(i - 1, j)
            mx2, mn2  = fn(i, j - 1)
            mx, mn = max(mx1, mx2) * grid[i][j], min(mn1, mn2) * grid[i][j]
            return (mx, mn) if grid[i][j] > 0 else (mn, mx)
        mx, _ = fn(m - 1, n - 1)
        return -1 if mx < 0 else mx % 1_000_000_007

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