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 <stdio.h>
#include <vector>
#include <iostream>
#include <queue>
#include <bits/stdc++.h>
 
#define rep(i,n) for (int i=0;i<(n);++i)
using namespace std;
typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<double> vd;
typedef pair<int, int> pi;


int main() {
  ll n;
  cin >> n;
  if (n % 2 == 1) {
    cout << 0 << endl;
    return 0;
  }

  ll ans = 0;
  ll m = 10;  
  while(n / m){
    ans += n / m;
    m *= 5;
  }
  cout << ans << endl;
  return 0;
}