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;
bool prime(long long n) {
  if (n == 1) return false;
  if (n % 2 == 0) return false;
  long long i;
  for (i = 3; i * i <= n; i += 2)
    if (n % i == 0) return false;
  return true;
}
long long primeDivisor(long long n) {
  {
    for (long long i = 3; i * i <= n; i += 2) {
      if (prime(i))
        if (n % i == 0) return i;
    }
  }
  return 1;
}
int main() {
  long long n, d, cnt = 0;
  bool w = false;
  cin >> n;
  if (prime(n)) {
    cout << "1" << endl;
    return 0;
  }
  if (n % 2 == 0)
    cnt = n / 2;
  else {
    while (n != 0) {
      d = primeDivisor(n);
      cnt++;
      n -= d;
      if (n % 2 == 0) {
        w = true;
        cnt += n / 2;
      }
      if (w) break;
    }
  }
  cout << cnt << endl;
  return 0;
}
