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 <bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i=0;i<n;i++)
#define rep1(i,n) for(int i=1;i<n+1;i++)
#define sort(A) sort(A.begin(),A.end())
#define reverse(A) reverse(A.begin(),A.end())
typedef long long ll;
typedef pair<int,int> P;

int main(){
  
  ll n;
  cin >>n;
  if(n%2 == 1){
    cout << 0 << endl;
    return 0;
  }
  ll ans = 0;
  for(int i=1;pow(5,i)<=n;i++){
    ans += n/(long long)(pow(5.0f,i)*2);
  }
  cout << ans << endl;
}