Let f(x) be the number of zeroes at the end of x!. Recall that x! = 1 2 3 ... x and by convention, 0! = 1.

For example, f(3) = 0 because 3! = 6 has no zeroes at the end, while f(11) = 2 because 11! = 39916800 has two zeroes at the end.

Given an integer k, return the number of non-negative integers x have the property that f(x) = k.

  Example 1:

Input: k = 0 Output: 5 Explanation: 0!, 1!, 2!, 3!, and 4! end with k = 0 zeroes.

Example 2:

Input: k = 5 Output: 0 Explanation: There is no x such that x! ends in k = 5 zeroes.

Example 3:

Input: k = 3 Output: 5

  Constraints:

0 <= k <= 10⁹