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>
#pragma GCC optimize "trapv"
const long long int m = 1073741824;
using namespace std;
long long int N = 1e6;
bool isprime(long long int x) {
  bool prime[N + 1];
  memset(prime, true, sizeof(prime));
  for (long long int p = 2; p * p <= N; p++) {
    if (prime[p] == true) {
      for (long long int i = p * p; i <= N; i += p) {
        prime[i] = false;
      }
    }
  }
  if (prime[x])
    return true;
  else
    return false;
}
bool isperfect(long long int n) {
  long double k = sqrt(n);
  return (k - (floor(k)) == 0);
}
int32_t main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  ;
  long long int n;
  cin >> n;
  if (n & 1) {
    if (isperfect(n)) {
      long long int k = sqrt(n);
      n -= k;
      cout << n << endl;
      if (n % 2 == 0) {
        cout << (n / 2) + 1;
      }
    } else
      cout << (1 + (n - 3) / 2);
  } else {
    long long int d = n / 2;
    cout << d;
  }
  return 0;
}
