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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) {
    long long n;
    cin >> n;
    vector<long long> v;
    long long tn = n;
    for (long long i = 2; i * i <= tn; i++) {
      if (n % i == 0) {
        v.push_back(i);
        while (n % i == 0) n = n / i;
      }
    }
    if (n > 1) v.push_back(n);
    sort(v.begin(), v.end());
    if (tn % 2 == 0)
      cout << tn / v[0];
    else {
      cout << 1 + (tn - v[0]) / 2;
    }
  }
}
