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: Wrong Answer)

// includes {{{
#include<iostream>
#include<iomanip>
#include<algorithm>
#include<vector>
#include<stack>
#include<queue>
#include<map>
#include<set>
#include<tuple>
#include<cmath>
#include<random>
#include<cassert>
#include<bitset>
#include<cstdlib>
// #include<deque>
// #include<multiset>
// #include<cstring>
// #include<bits/stdc++.h>
// }}}
using namespace std;
using ll = long long;

int main() {
  std::ios::sync_with_stdio(false), std::cin.tie(0);
  ll n;
  cin >> n;
  if(n % 2) {
    ll x = 5;
    ll ans = 0;
    while(n / x - n / x / 2) ans += n / x - n / x / 2, x *= 5;
    cout << ans << endl;
  } else {
    ll x = 5;
    ll ans = 0;
    while(n / x / 2) ans += n / x / 2, x *= 5;
    cout << ans << endl;
  }
  return 0;
}