Problem:
Score : 500 points

Problem Statement

For an integer n not less than 0, let us define f(n) as follows:

f(n) = 1 (if n < 2)

f(n) = n f(n-2) (if n \geq 2)

Given is an integer N. Find the number of trailing zeros in the decimal notation of f(N).

Constraints

0 \leq N \leq 10^{18}

Input

Input is given from Standard Input in the following format:

N

Output

Print the number of trailing zeros in the decimal notation of f(N).

Sample Input 1

12

Sample Output 1

1

f(12) = 12 × 10 × 8 × 6 × 4 × 2 = 46080, which has one trailing zero.

Sample Input 2

5

Sample Output 2

0

f(5) = 5 × 3 × 1 = 15, which has no trailing zeros.

Sample Input 3

1000000000000000000

Sample Output 3

124999999999999995

Solution in C (GCC 5.4.1) (judge verdict: Runtime Error)

#include<stdio.h>
long t(long a)
{
	if (a % 10 != 0) {
		return t(a - 2);
	}
	else {
		return a;
	}
}

int countz(long a)
{
	int count;
	while (a % 10 == 0) {
		count++;
	}
	return count;
}

long numz(long a)
{
	long A=t(a);
	if(a>=11){
		return countz(A) * numz(A - 10);
	}else {
		return 1;
	}
}

int main(void)
{
	long a;

	scanf("%ld", &a);

	if (a % 2 == 1) {
		printf("0");
		return;
	}

	printf("%ld",numz(a));

}