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() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n;
  cin >> n;
  std::vector<std::pair<long long, long long>> factors;
  long long t = n;
  for (long long i = 2; i <= sqrt(t); i++) {
    bool inserted = false;
    while (n % i == 0) {
      n /= i;
      if (!inserted) {
        factors.push_back(std::make_pair(i, 1));
        inserted = true;
      } else
        factors[factors.size() - 1].second++;
    }
  }
  bool counted = false;
  long long ans = 0;
  if (n != 1) {
    ans++;
  }
  if (!counted) {
    if ((int)factors.size() == 0) {
      cout << "d"
           << "\n";
      return 0;
    }
    for (int i = 0; i < factors.size(); i++) {
      ans += factors[i].second;
    }
    cout << ans << "\n";
  }
  return 0;
}
