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;
inline int rit() {
  int f = 0, key = 1;
  char ch;
  do {
    ch = getchar();
    if (ch == '-') key = -1;
  } while (ch < '0' || ch > '9');
  do {
    f = f * 10 + ch - '0';
    ch = getchar();
  } while (ch >= '0' && ch <= '9');
  return f * key;
}
long long n;
int32_t main(void) {
  scanf("%lld", &n);
  if (n & 1) {
    bool fd = false;
    for (long long i = 2; i * i <= n; ++i) {
      if (n % i == 0) {
        n -= i;
        fd = true;
        break;
      }
    }
    if (fd)
      printf("%lld\n", n / 2 + 1);
    else
      printf("1\n");
  } else {
    printf("%lld\n", n / 2);
  }
  return 0;
}
