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>
#define int long long
#define pb push_back
#define mp make_pair
using namespace std;
const int inf = 1e18 + 7;
const int N = 1e6 + 7;
const int md = 1e9 + 7;
int tz(int n) 
{ 
    int cnt = 0; 
    for (int i = 5; n / i >= 1; i *= 5) 
        cnt += n / (i); 
  
    return cnt; 
} 
void sol() {
    int n;
    cin >> n;
    if(n % 2) {
        cout << 0;
        return;
    }
    n /= 2;
    cout << tz(n);
    
    //cout << (n % 2 == 0) ? tz(n) : 0;
}

int32_t main() 
{
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int T = 1;
  //cin >> T; while(T--) 
  sol();
  return 0;
}