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>
int look(long long h) {
  long long j, m = 1;
  for (j = 2; j <= sqrt(h); j++) {
    if (h % j == 0) {
      m = 0;
      break;
    }
  }
  return m;
}
int main() {
  long long a, num;
  long long i, j, m;
  while (~scanf("%lld", &a) && a) {
    if (look(a))
      printf("1\n");
    else if (a % 2 == 0)
      printf("%lld\n", a / 2);
    else {
      for (i = 2; i <= a; i++) {
        m = 0;
        if (a % i == 0) {
          m = look(i);
          if (m) {
            printf("%lld\n", 1 + (a - i) / 2);
          }
        }
        if (m == 1) break;
      }
    }
  }
  return 0;
}
