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++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include<bits/stdc++.h>
using namespace std;
#define ll long long int
ll recur(ll N)
{
  if(N==0 || N==1)
    return 1;
  return N*recur(N-2);
}
int main(){
ios::sync_with_stdio(0);
  cin.tie(0);
  ll n,cnt=0;
  scanf("%ld",&n);
  ll ans = recur(n);
  while(ans%10==0)
  {
    ++cnt;
    ans%=10;
  }
  printf("%ld\n",cnt);
  return 0;
}