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(){
  long long n;
  long long k = 0;
  long long count = 0;
  
  scanf("%lld", &n);
  
  if(n % 2 == 1){
    printf("0\n");
  }
  else{
    for(long long i = n; i > 9; i -= 2){
      if(i % 10 == 0){
        long long m = i;
        while((m % 10) == 0){
          m /= 10;
          k++;
        }
      }
    }   
     printf("%lld\n", k);
  }
 
  return 0;
}