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<long long> v;
bool prime[100005];
void sieve() {
  memset(prime, true, sizeof(prime));
  for (int i = 4; i <= 100000; i += 2) {
    prime[i] = false;
  }
  prime[1] = prime[0] = false;
  for (int i = 3; i * i <= 100000; i += 2) {
    if (prime[i]) {
      for (int j = i * i; j <= 100000; j = j + (2 * i)) {
        prime[j] = false;
      }
    }
  }
  v.push_back(2);
  for (int i = 3; i <= 100000; i += 2) {
    if (prime[i]) {
      v.push_back(i);
    }
  }
}
int main() {
  long long n, s = 0, l = 0;
  sieve();
  cin >> n;
  for (long long i = 0; i < v.size(); i++) {
    if (n % 2 == 0) {
      s = s + (n / 2);
      l = 1;
      break;
    } else if (n % v[i] == 0) {
      s++;
      n = n - v[i];
      if (n == 0) {
        break;
      }
    }
  }
  if (l == 0) {
    s = 1;
  }
  cout << s;
  return 0;
}
