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>
const long long INF = 1e18L + 1;
const int IINF = 1e9 + 1;
using namespace std;
template <class TH>
void _dbg(const char *sdbg, TH h) {
  cerr << sdbg << '=' << h << endl;
}
template <class TH, class... TA>
void _dbg(const char *sdbg, TH h, TA... a) {
  while (*sdbg != ',') cerr << *sdbg++;
  cerr << '=' << h << ',';
  _dbg(sdbg + 1, a...);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long n;
  cin >> n;
  if ((n & 1) == 0) {
    cout << n / 2 << "\n";
  } else {
    for (long long i = 3; i * i <= n; i++) {
      if (n % i == 0) {
        cout << (n - i) / 2 + 1 << "\n";
        return 0;
      }
    }
    cout << 1 << "\n";
  }
  return 0;
}
