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;
int main() {
  int n;
  int stej = 0;
  int primes[100004] = {0, 0};
  int hehe = 2;
  while (hehe < 100004) {
    primes[hehe] = hehe;
    hehe++;
  }
  for (int i = 2; i < 100004; i++) {
    if (primes[i != 0]) {
      for (int k = i * i - 1; k < 100004; k++) {
        if (primes[k] != 0) {
          if (k % i == 0) {
            primes[k] = 0;
          } else {
            primes[k] = k;
          }
        }
      }
    }
  }
  cin >> n;
  for (int j = 2; j < 100004; j++) {
    if (primes[j] != 0) {
      if (n % j == 0) {
        n = n - j;
        if (n == 0) {
          cout << "1" << endl;
          return 0;
        }
        stej++;
        break;
      }
    }
  }
  stej = stej + n / 2;
  cout << stej << endl;
  return 0;
}
