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;
bool mark[100000 + 100];
unsigned long long prime[100000 + 100];
int i, j, k;
void sieve() {
  int lmt = sqrt(1 + 100000 + 100);
  for (i = 3; i <= lmt; i += 2) {
    if (!(mark[i])) {
      for (j = i * i; j <= 100000 + 100; j += 2 * i) {
        mark[j] = 1;
      }
    }
  }
}
void gen_prime_upto(int LMT) {
  prime[1] = 2;
  for (i = 3, j = 2; i <= 100000 + 100; i += 2) {
    if (!(mark[i])) prime[j++] = i;
  }
}
bool is_prime(unsigned long long n) {
  int lmt = sqrt(1 + n);
  for (k = 2; k <= lmt; k++) {
    if (n % k == 0) return 0;
  }
  return 1;
}
int main() {
  sieve();
  gen_prime_upto(100000 + 100 + 100);
  unsigned long long n;
  scanf("%lld", &n);
  unsigned long long cnt = 0;
  if (n % 2 == 0)
    cnt = n / 2;
  else if (is_prime(n) == 1)
    cnt = 1;
  else {
    unsigned long long lmt = sqrt(1 + n);
    for (i = 2; prime[i] <= lmt; i++) {
      if (n % prime[i] == 0) {
        n -= prime[i];
        break;
      }
    }
    cnt = n / 2 + 1;
  }
  printf("%lld\n", cnt);
}
