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 long long mod = 1e9 + 7;
long double PI = acosl(-1);
const long long inf = 1e15;
const int nmax = 1e6 + 5;
bool iscomposit[nmax];
void siv() {
  iscomposit[1] = true;
  for (int i = 2; i < nmax; i++) {
    if (!iscomposit[i]) {
      for (int j = i + i; j < nmax; j += i) {
        iscomposit[j] = true;
      }
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  siv();
  long long n;
  cin >> n;
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0 and iscomposit[i] == false) {
      long long ans = 1;
      n -= i;
      ans += n / 2;
      cout << ans << endl;
      exit(0);
    }
  }
  cout << "1" << endl;
}
