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: Accepted)

// E

#include <bits/stdc++.h>
using ll = long long;
using namespace std;
ll mod=1e9+7;

int main(){
  ll n;
  cin>>n;
  ll cnt(0);
  if (n%2==1){
    cout << 0 << endl;
  }else{
    ll tmp = 1;
    while(tmp<n){
      tmp*= 5;
      cnt += n / (tmp*2);
    }
    cout <<cnt << endl;
  }
  return 0;
}