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;
vector<unsigned long long> generate_primes(unsigned long long max_elem) {
  vector<unsigned long long> primes;
  bool b;
  for (unsigned long long i = 2; i <= (max_elem); i++) {
    b = true;
    for (unsigned long long prime : primes) {
      if (prime > sqrt(i)) {
        break;
      }
      if (i % prime == 0) {
        b = false;
        break;
      }
    }
    if (b) {
      primes.push_back(i);
    }
  }
  return primes;
}
int main(void) {
  long long n;
  cin >> n;
  vector<unsigned long long> primes = generate_primes((int)(sqrt(n)));
  long long p = n;
  for (auto e : primes) {
    if (n % e == 0) {
      p = e;
      break;
    }
  }
  cout << (n / p);
}
