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;
long long n;
bool vis[100005];
bool ck_p(long long n) {
  for (long long i = 2; i <= sqrt(n); i++)
    if (n % i == 0) return false;
  return true;
}
void find_p() {
  for (int i = 2; i <= 100005; i++) {
    if (vis[i] == 1) continue;
    int cnt = 2;
    if (ck_p(i)) {
      while (i * cnt <= 100005) {
        vis[i * cnt] = 1;
        cnt++;
      }
    }
  }
}
int main() {
  find_p();
  cin >> n;
  if (ck_p(n)) {
    cout << 1 << endl;
  } else {
    if (n % 2 == 0) {
      cout << n / 2 << endl;
    } else {
      for (int i = 2; i <= sqrt(n); i++) {
        if (!vis[i] && n % i == 0) {
          n -= i;
        }
      }
      cout << 1 + n / 2 << endl;
    }
  }
  return 0;
}
