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: Accepted)

#include <stdio.h>
#include <math.h>

int main(void){
	long long int n;
	long long int ans=0;
	scanf("%lld", &n);
	
	if(n%2==1)	printf("0");
	else{
		for(int i=1; i <= (int)(log10l((long double)n/(long double)2.0)/log10l(5.0));i++){
			ans += (n-n%(long long int)(pow(5,i)*2) ) / (long long int)(pow(5,i)*2);
		}
      printf("%lld", ans);
	}
	return 0;
}