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 arr[100006];
set<long long> s;
void make() {
  arr[0] = arr[1] = 1;
  for (long long i = 2; i < 100005; i++) {
    if (arr[i] == 0) {
      long long t = 2;
      while (t * i <= 100005) {
        arr[t * i] = 1;
        t++;
      }
    }
  }
  for (long long i = 0; i < 100005; i++) {
    if (arr[i] == 0) {
      s.insert(i);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  make();
  long long n;
  cin >> n;
  long long ans = 0;
  while (n != 0) {
    for (set<long long>::iterator it = s.begin(); it != s.end(); it++) {
      if (n % (*it) == 0) {
        n -= *it;
        ans++;
        ans += (n / 2);
        cout << ans;
        return 0;
      }
    }
    cout << 1;
    return 0;
  }
  cout << ans;
  return 0;
}
