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;
long long n, divi;
void sieve(long long ub) {
  vector<bool> is_prime(ub + 1, true);
  is_prime[0] = is_prime[1] = false;
  for (long long i = 2; i <= ub; i++) {
    if (is_prime[i]) {
      if (n % i == 0) {
        divi = i;
        return;
      }
      for (int j = i * 2; j <= ub; j += i) is_prime[j] = false;
    }
  }
}
bool isprime(long long x) {
  for (long long i = 2; i * i <= x; i++)
    if (n % i == 0) return false;
  return true;
}
int main() {
  cin >> n;
  if (n % 2 == 0)
    cout << (n / 2) << endl;
  else if (isprime(n))
    cout << 1 << endl;
  else {
    sieve((long long)sqrt(n));
    cout << (n / divi) << endl;
  }
}
