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<stdlib.h>
#include<string.h>

typedef long long int int64;

int main()
{
    int64 a,b=0;
    int64 arr[30]={0};
    int64 ans=0;
    int i=0,j;

    scanf("%lld",&a);

    b = a;
    //printf("%lld\n",b);

    if(b%2==1)ans=0;
    else
    {
        while(b>0)
        {
            //printf("b = %lld\n",b);
            b/=5;
            arr[i]=b;
            i++;
        }
        for(j=0;j<i;j++)
        {
            ans+=(arr[j]/2);
        }
    }
    printf("%lld",ans);

}