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: Time Limit Exceeded)

#include<stdio.h>
int main(void)
{
  long int n,i,j,a,b,c;
  scanf("%ld",&n);
  if(n%2==1){
    printf("0\n");
  }
  a=0;
  if(n%2==0){
    for(i=0;i<5;i++){
      if((n-2*i)%10==0){
        c=n-2*i;
        break;
      }
      if((n-2*i)%10!=0){
        continue;
      }
    }
    /*printf("%ld\n",c);*/
    for(i=0;i<n/10;i++){
      b=c;
      for(j=0;j<19;j++){
        if(b%10!=0){
          break;
        }
        if(b%10==0){
          a++;
          b=b/10;
        }
      }
      c=c-10;
    }
  printf("%ld\n",a);
  }
  return 0;
}