Double Factorial

For a positive integer $n$, its factorial $n!$ is the product of all integers from $1$ to $n$. Define the double factorial of $n$ as the product of the first $n$ factorials:

$1! \cdot 2! \cdot 3! \cdots n!$

Given $n$, determine the number of trailing zeros in the decimal representation of this double factorial.

A single line contains one integer $n$ ($1 \le n \le 10^{18}$).

Print a single line with the number of trailing zeros of the double factorial of $n$.

For $n = 11$, the double factorial equals $265790267296391946810949632000000000$, which has $9$ trailing zeros.