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)

#include<iostream>
#include<algorithm>
#include<cmath>
#include<map>
#include<string>
#include<set>
#include<vector>
using namespace std;

template<typename T>
T powll(T base, T exp){
  T ret = 1;
  if(exp == 0) return 1;
  while(exp != 1){
    if((int)exp&1){
      ret *= base;
      exp--;
    }
    exp/=2;
    base *= base;
  }
  ret *= base;
  return ret;
}

int solve(){
  long long n;  cin >> n;
  if(n&1){
    cout << 0 << endl;
    return 0;
  }
  string s=to_string(n);
  int sn = s.length();
  long long ans = 0;
  ans += n/10ll;
  for(long long i=5; i*10<=n; i*=5){
    ans += n/(10ll * i);
  }
  cout << ans << endl;
}

int main(){
  solve();
  return 0;
}