Given an integer n, return the number of trailing zeroes in n!.

Note that n! = n (n - 1) (n - 2) ... 3 2 1.

  Example 1:

Input: n = 3 Output: 0 Explanation: 3! = 6, no trailing zero.

Example 2:

Input: n = 5 Output: 1 Explanation: 5! = 120, one trailing zero.

Example 3:

Input: n = 0 Output: 0

  Constraints:

0 <= n <= 10⁴

  Follow up: Could you write a solution that works in logarithmic time complexity?