Factorials with an even number of trailing zeroes

The factorial of a positive integer $n$ is written $n!$ and is defined like this.

$n! = 1 \times 2 \times 3 \times 4 \times \cdots \times (n-1) \times n$

The value of $0!$ is taken to be $1$. As $n$ grows, $n!$ grows very quickly. Here are a few values.

• $0! = 1$
• $1! = 1$
• $2! = 2$
• $3! = 6$
• $4! = 24$
• $5! = 120$
• $10! = 3628800$
• $14! = 87178291200$
• $18! = 6402373705728000$
• $22! = 1124000727777607680000$

For some $n$ the number of trailing zeroes of $n!$ is odd. That happens for $5!$ and $18!$. For other $n$ it is even, as with $0!$, $10!$ and $22!$.

Given $n$, count how many of $0!, 1!, 2!, 3!, \ldots, (n-1)!, n!$ have an even number of trailing zeroes.

Standard input holds one query per line, at most $1000$ of them. Each query is a single integer $n$ ($0 \le n \le 10^{18}$).

The last line holds $-1$. That line is not a query, it only marks the end of the input.

Print one line per query. The line holds how many of $0!, 1!, 2!, 3!, \ldots, n!$ have an even number of trailing zeroes.

Print nothing for the $-1$ that ends the input.