Factorials With Exactly K Trailing Zeros

Hard

The factorial is x! = 1 · 2 · … · x, with 0! = 1. For instance, 10! = 3628800 ends in two zeros. Given zeros, count how many non-negative integers x have x! ending in exactly zeros zeros in decimal. For large zeros you cannot compute the factorials themselves, so look for the pattern.

Example 1

zeros=1 → 5

Exactly one trailing zero appears in 5!, 6!, 7!, 8! and 9! (5! = 120, while 10! already ends in 00). The answer is 5.

Example 2

zeros=5 → 0

24! ends in 4 zeros and 25! jumps straight to 6, because the factor 25 adds two at once. Exactly five never happens, so the answer is 0.

Example 3

zeros=0 → 5

Constraints

0 ≤ zeros ≤ 10⁹

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Input

zeros=1

Expected

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