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`