You are given an integer number n. The following algorithm is applied to it:

  1. if n = 0, then end algorithm; 
  2. find the smallest prime divisor d of n; 
  3. subtract d from n and go to step 1. 



Determine the number of subtrations the algorithm will make.

Input

The only line contains a single integer n (2 ≤ n ≤ 10^{10}).

Output

Print a single integer — the number of subtractions the algorithm will make.

Examples

Input


5


Output


1


Input


4


Output


2

Note

In the first example 5 is the smallest prime divisor, thus it gets subtracted right away to make a 0.

In the second example 2 is the smallest prime divisor at both steps.
#include <bits/stdc++.h>
using namespace std;
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
inline void Routine() {
  srand(time(NULL));
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
template <class T>
istream &operator>>(istream &in, vector<T> &v) {
  for (auto &i : v) {
    in >> i;
  }
  return in;
}
template <class T>
ostream &operator<<(ostream &out, const vector<T> &v) {
  for (auto &i : v) {
    out << i << ' ';
  }
  return out;
}
int main() {
  Routine();
  long long n, t, co = 0, ans = 0, x = 0, y = 0, b, k;
  cin >> n;
  if (n % 2 == 0) {
    cout << n / 2;
    return 0;
  } else {
    for (long long i = 2; i <= n; i++) {
      if (n % i == 0) {
        co = i;
        break;
      }
    }
    cout << n / co;
  }
  return 0;
}
