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() {
  long long n, i, j, k, l, e;
  scanf("%lld", &n);
  long long a[100005], p[100000];
  for (i = 2; i <= 100000; i++) {
    a[i] = 1;
  }
  for (i = 2; i <= 1000; i++) {
    for (j = 2 * i; j <= 100000; j += i) {
      a[j] = 0;
    }
  }
  j = 0;
  for (i = 3; i <= 100000; i++) {
    if (a[i] == 1) {
      p[j] = i;
      j++;
    }
  }
  p[j] = 1000000;
  if (n % 2 == 0)
    printf("%lld\n", n / 2);
  else {
    i = 0;
    e = 0;
    while (p[i] <= sqrt(n)) {
      if (n % p[j] == 0) {
        n -= p[j];
        e = 1;
        break;
      } else
        i++;
    }
    if (e == 1)
      printf("%lld\n", 1 + n / 2);
    else
      printf("1\n");
  }
  return 0;
}
