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 const mod = 99824435;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long t, n, i, j, k = 0, cnt = 0, sum1 = 0, ccnt = 0, x = -1, y, z, m,
                        sum = 0, x1, y1 = 0, z1, d, total = 0;
  cin >> n;
  if (n == 2 || n == 3 || n == 5 || n == 7) {
    cout << 1 << endl;
  } else {
    for (i = 2; i * i <= n; i++) {
      if (n % i == 0) {
        x = i;
        break;
      }
    }
    if (x == -1) {
      cout << 1 << endl;
    } else {
      if (((n - x) + 1) % 2 == 0) {
        cout << ((n - x) + 1) / 2 << endl;
      } else {
        cout << (((n - x) + 1) / 2) + 1 << endl;
      }
    }
  }
  return 0;
}
