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 dx[] = {-1, 1, 0, 0};
long long dy[] = {0, 0, -1, 1};
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n;
  cin >> n;
  long long cnt = 0;
  while (true) {
    if (!(n & 1)) {
      cout << cnt + n / 2;
      return 0;
    }
    bool stop = false;
    for (long long i = 3; i * i <= n; i += 2) {
      if (n % i == 0) {
        n -= i;
        stop = 1;
        cnt++;
        break;
      }
    }
    if (!stop) {
      cout << 1 << "\n";
      return 0;
    }
  }
}
