<p>You are given a 2D integer array <code>grid</code> of size <code>m x n</code>, where each cell contains a positive integer.</p>

<p>A <strong>cornered path</strong> is defined as a set of adjacent cells with <strong>at most</strong> one turn. More specifically, the path should exclusively move either <strong>horizontally</strong> or <strong>vertically</strong> up to the turn (if there is one), without returning to a previously visited cell. After the turn, the path will then move exclusively in the <strong>alternate</strong> direction: move vertically if it moved horizontally, and vice versa, also without returning to a previously visited cell.</p>

<p>The <strong>product</strong> of a path is defined as the product of all the values in the path.</p>

<p>Return <em>the <strong>maximum</strong> number of <strong>trailing zeros</strong> in the product of a cornered path found in </em><code>grid</code>.</p>

<p>Note:</p>

<ul>
	<li><strong>Horizontal</strong> movement means moving in either the left or right direction.</li>
	<li><strong>Vertical</strong> movement means moving in either the up or down direction.</li>
</ul>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/03/23/ex1new2.jpg" style="width: 577px; height: 190px;" />
<pre>
<strong>Input:</strong> grid = [[23,17,15,3,20],[8,1,20,27,11],[9,4,6,2,21],[40,9,1,10,6],[22,7,4,5,3]]
<strong>Output:</strong> 3
<strong>Explanation:</strong> The grid on the left shows a valid cornered path.
It has a product of 15 * 20 * 6 * 1 * 10 = 18000 which has 3 trailing zeros.
It can be shown that this is the maximum trailing zeros in the product of a cornered path.

The grid in the middle is not a cornered path as it has more than one turn.
The grid on the right is not a cornered path as it requires a return to a previously visited cell.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/03/25/ex2.jpg" style="width: 150px; height: 157px;" />
<pre>
<strong>Input:</strong> grid = [[4,3,2],[7,6,1],[8,8,8]]
<strong>Output:</strong> 0
<strong>Explanation:</strong> The grid is shown in the figure above.
There are no cornered paths in the grid that result in a product with a trailing zero.
</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;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= m * n &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= grid[i][j] &lt;= 1000</code></li>
</ul>

class Solution:
    def maxTrailingZeros(self, grid: List[List[int]]) -> int:
        def exponents(n):
            r = [0, 0]
            ns = [2, 5]
            for i in range(2):
                while n % ns[i] == 0:
                    n //= ns[i]
                    r[i] += 1
            return r
        m, n = len(grid), len(grid[0])
        horizontal = [[None] * n for i in range(m)]
        vertical = [[None] * n for i in range(m)]
        for i in range(m):
            for j in range(n):
                horizontal[i][j] = exponents(grid[i][j])
                vertical[i][j] = exponents(grid[i][j])
                if j > 0:
                    horizontal[i][j][0] += horizontal[i][j - 1][0]
                    horizontal[i][j][1] += horizontal[i][j - 1][1]
                if i > 0:
                    vertical[i][j][0] += vertical[i - 1][j][0]
                    vertical[i][j][1] += vertical[i - 1][j][1]

        res = 0
        for i in range(m):
            for j in range(n):
                g = [horizontal[i][j][0], horizontal[i][j][1]]
                h = [[horizontal[i][j][0], horizontal[i][j][1]], [horizontal[i][-1][0], horizontal[i][-1][1]]]
                if j > 0:
                    h[-1][0] -= horizontal[i][j - 1][0]
                    h[-1][1] -= horizontal[i][j - 1][1]
                    g[0] -= horizontal[i][j - 1][0]
                    g[1] -= horizontal[i][j - 1][1]
                v = [[vertical[i][j][0], vertical[i][j][1]], [vertical[-1][j][0], vertical[-1][j][1]]]
                if i > 0:
                    v[-1][0] -= vertical[i - 1][j][0]
                    v[-1][1] -= vertical[i - 1][j][1]
                for k in range(len(h)):
                    for l in range(len(v)):
                        res = max(res, min(h[k][0] + v[l][0] - g[0], h[k][1] + v[l][1] - g[1] ))
        return res
72900