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 long long int mod = 998244353;
bool comp(pair<long long int, long long int> p1,
          pair<long long int, long long int> p2) {
  if (p1.first == p2.first) return p1.second < p2.second;
  return p1.first < p2.first;
}
bool isprime(long long int n) {
  for (long long int i = 2; i * i <= n; i++) {
    if (n % i == 0) return false;
  }
  return true;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int n;
  cin >> n;
  if (n % 2 == 0) {
    cout << n / 2;
    return 0;
  } else if (isprime(n)) {
    cout << 1;
    return 0;
  }
  bool prime[1000000];
  memset(prime, 1, sizeof(prime));
  for (long long int i = 2; i < 1000000; i++) {
    if (prime[i] == true) {
      for (long long int k = i * i; k < 1000000; k += i) {
        prime[k] = false;
      }
    }
  }
  for (long long int i = 3; i * i <= n; i++) {
    if (prime[i] && n % i == 0) {
      n -= i;
      long long int ans = n / 2;
      cout << ans + 1;
      return 0;
    }
  }
}
