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, k, b, c, m, ans;
vector<long long> v;
int main() {
  cin >> n;
  k = floor(sqrt(double(n))) + 3;
  bool f = true;
  if (n % 2 == 0) {
    ans = n / 2;
  } else {
    for (int i = 2; i < k; i++) {
      if (n % i == 0) {
        f = false;
        c = i;
        break;
      }
    }
    if (f) {
      ans = 1;
    } else {
      ans++;
      n -= c;
      ans += n / 2;
    }
  }
  cout << ans;
  return 0;
}
