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>
#pragma GCC optimize "trapv"
const long long int m = 1073741824;
using namespace std;
long long int smallPrime(long long int n) {
  if (n % 2 == 0)
    return 2;
  else {
    for (long long int i = 3; i * i <= n; i += 2) {
      if (n % i == 0) return i;
    }
  }
  return n;
}
int32_t main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  ;
  long long int n;
  cin >> n;
  if (n & 1) {
    long long int k = smallPrime(n);
    long long int ans = n - k;
    cout << 1 + ans / 2;
  } else {
    long long int d = n / 2;
    cout << d;
  }
  return 0;
}
