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

int main(void){
  char str[30];
  long int x,i,N,sum;
  
  fgets(str,sizeof(str),stdin);
  N=atol(str);
  
  if(N%2==1){
    printf("0");
  }else{
    N=N/2;
    sum=0;
    for(i=1;(long int)powl(5,i)<=N;i++){
      x=(long int)powl(5,i);
      sum+=N/x;
    }
    printf("%ld\n",sum);
  }
  return 0;
}