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

int main(){
  long n,i,now,r,mod;

  scanf("%ld", &n);

  i = n % 2;
  now = 1;
  r=0;
  while(i<n)
  {
  //printf("i: %ld, now: %ld, r: %ld\n", i, now, r);
    if (i != 0) now *= i;
    i += 2;

   // printf("i: %ld, now: %ld, r: %ld\n", i, now, r);
   // printf("now  per 10: %d\n", now % 10);
    while ((now % 10) == 0)
    {
    //printf("now: %ld\n",now);
    if (now==0) break;
      r++;
      now /= 10;
    }
  }
  printf("%ld\n",r);

  return 0;
}