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;
const int base = 1000 * 1000 * 1000;
const int N = (int)2e5 + 2;
const int INF = (int)1e9 + 7;
const long double pi = acos(-1);
const long long linf = (long long)1e18;
long long n;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n;
  long long ans = 0, last = n;
  if (n & 1) {
    long long mn = linf;
    for (long long div = 2; div <= sqrt(n); ++div) {
      if (n % div == 0) {
        mn = min(mn, div);
        while (n % div == 0) {
          n /= div;
        }
      }
    }
    if (n > 1) {
      mn = min(mn, n);
    }
    last -= mn;
    ans++;
  }
  ans += last / 2;
  cout << ans << "\n";
  return 0;
}
