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;
void fast() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
}
bool isprime(long long n) {
  long long i;
  for (i = 2; i * i < n; i++) {
    if (n % i == 0) return false;
  }
  if (i * i == n) return false;
  return true;
}
vector<bool> s(1e5 + 1, 1);
vector<long long> primes;
void seive() {
  s[0] = 0;
  s[1] = 0;
  for (int i = 2; i * i <= 1e5; i++) {
    if (s[i]) {
      for (int j = i * i; j <= 1e5; j += i) {
        s[j] = false;
      }
    }
  }
  for (int i = 2; i <= 1e5; i++) {
    if (s[i]) primes.push_back(i);
  }
}
void solve() {
  long long n;
  cin >> n;
  if (n % 2 == 0) {
    cout << n / 2 << '\n';
  } else if (isprime(n)) {
    cout << 1 << '\n';
  } else {
    seive();
    for (int i = 0; i < primes.size(); i++) {
      if (n % primes[i] == 0) {
        n -= primes[i];
        break;
      }
    }
    cout << n / 2 + 1 << '\n';
  }
}
int main() {
  fast();
  solve();
  return 0;
}
