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>
 #include<stdlib.h>
long long int fact(long long int);

	int main()
	{
   	 long long int n,a,m,l=0,i,t;
   	 t=1000000007;
    	scanf("%lld",&n);
    	a=fact(n);
    	//printf("%lld",fact(n));
    	for(i=5;a/i>=1;i*=5)
    	{
    	    l+=n/i;
    	}
    	printf("%lld",l-1);
    	return 0;
	}

long long int fact(long long int n)
	{
	    long long int t;
	    t=1000000007;
    	while(n>=1)
    	{
        	return (n*fact(n-2))%t;
    	}
    	return 1;
	}