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;
const int maxn = 1e6;
int prime[maxn];
bool isprime[maxn + 10];
void sieve(int n) {
  int i, j;
  isprime[0] = isprime[1] = false;
  for (i = 2; i * i <= n; i++) {
    if (isprime[i]) {
      for (j = i * i; j <= n; j += i) {
        isprime[j] = false;
      }
    }
  }
}
long long ans = 0;
void factorize(long long n) {
  long long idx = 0, pf = prime[idx];
  while (pf * pf <= n) {
    if (n % pf == 0) {
      long long loop = n / pf;
      n -= (loop * pf);
      ans += loop;
    }
  }
}
int main() {
  memset(isprime, true, sizeof isprime);
  sieve(maxn);
  int cnt = 0;
  int i, j;
  for (i = 0; i < maxn; i++) {
    if (isprime[i]) {
      prime[cnt] = i;
      cnt++;
    }
  }
  long long n;
  scanf("%lld", &n);
  bool apakahPrima = true;
  for (i = 0; i < cnt; i++) {
    if (n % prime[i] == 0) {
      apakahPrima = false;
      n -= prime[i];
      ans++;
      break;
    }
  }
  if (apakahPrima) {
    printf("1\n");
  } else {
    long long loop = n / 2LL;
    ans += loop;
    printf("%lld\n", ans);
  }
  return 0;
};
