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 maxn = 10000000000 + 5;
long long n, minp;
long long ans;
int is_prime(long long x) {
  for (long long i = 2; i * i <= x; i++) {
    if (x % i == 0) {
      minp = i;
      return 0;
    }
  }
  return 1;
}
int solve() {
  while (n != 0) {
    if (is_prime(minp) && n % minp == 0) {
      ans += (n / minp);
      n = n % minp;
      minp = 2;
    } else {
      while (!is_prime(minp)) {
        minp++;
      }
    }
  }
  return 1;
}
int main() {
  cin >> n;
  ans = 0;
  if (is_prime(n))
    cout << 1 << endl;
  else {
    solve();
    cout << ans << endl;
  }
  return 0;
}
