<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:
        class BaseTup:
            def __init__(self, a=0, b=0):
                self.a, self.b = a, b

            @classmethod
            def from_num(cls, n): 
                c2 = c5 = 0
                while n and n%2==0:
                    c2 += 1
                    n = n>>1
                while n and n%5==0:
                    c5 += 1
                    n //= 5
                return BaseTup(c2, c5)

            @classmethod
            def sum(cls, iterable):
                return sum(iterable, start = BaseTup())

            def __add__(self, o):
                return BaseTup(self.a + o.a, self.b + o.b)

            def __sub__(self, o):
                return BaseTup(self.a - o.a, self.b - o.b)

            @property
            def trail_zeros(self):
                return min(self.a, self.b)

        
        m,n = len(grid), len(grid[0])
        ngrid = [[ BaseTup.from_num(grid[i][j]) for j in range(n)] for i in range(m)]

        up_sum = [BaseTup() for _ in range(n)]
        down_sum = [BaseTup.sum(ngrid[i][j] for i in range(m)) for j in range(n)]

        zmax = 0
        for i in range(m):
            rsum = BaseTup.sum(ngrid[i])
            lsum = BaseTup()
            for j in range(n):
                t = ngrid[i][j]
                lsum += t
                down_sum[j] -= t

                zmax = max(zmax, 
                    (lsum+up_sum[j]).trail_zeros,
                    (rsum+up_sum[j]).trail_zeros,
                    (lsum+down_sum[j]).trail_zeros,
                    (rsum+down_sum[j]).trail_zeros)

                rsum -= t
                up_sum[j] += t
                
        return zmax
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