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: Wrong Answer)

#include<stdio.h>
#include<math.h>
int main(void)
{
  long int n,i,j,a,b,c,p,d;
  scanf("%ld",&n);
  if(n%2==1){
    printf("0\n");
  }
  a=0;
  if(n%2==0){
    b=n;
    p=1;
    while(p*10<=n){
      d=b-(b%(10*p));
      a=a+(d/(10*p));
      printf("%ld\t",a);
      p=p*5;
      /*printf("%ld\n",p*10);*/
    }
  printf("%ld\n",a);
  }
  return 0;
}