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;
bool prime[100005];
void SieveOfEratosthenes() {
  memset(prime, true, sizeof(prime));
  for (long long p = 2; p * p < 100005; p++) {
    if (prime[p] == true) {
      for (long long i = p * p; i < 100005; i += p) prime[i] = false;
    }
  }
}
int main() {
  long long n;
  cin >> n;
  memset(prime, true, sizeof(prime));
  SieveOfEratosthenes();
  if (n % 2 == 0)
    cout << n / 2 << "\n";
  else {
    bool flag = 0;
    for (int p = 3; p <= 100004; p++) {
      if (prime[p])
        if (n % p == 0) {
          n = n - p;
          flag = 1;
          break;
        }
    }
    if (flag) {
      cout << n / 2 + 1 << "\n";
    } else {
      cout << "1"
           << "\n";
    }
  }
}
