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;
template <class T>
bool maxs(T& a, T& b) {
  if (a < b) {
    T t;
    t = a;
    a = b;
    b = t;
    return true;
  }
  return false;
}
template <class T>
bool maxi(T& a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
bool mini(T& a, T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n;
  cin >> n;
  long long cnt = 0;
  if (n & 1) {
    int f = 0;
    for (int i = 3; i <= sqrt(n); i++)
      if (n % i == 0) {
        f = 1;
        n -= i;
        break;
      }
    if (!f)
      ;
    n = 0;
    cnt++;
  }
  cout << cnt + n / 2;
  return 0;
}
