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 int N = 1e5 + 1;
const int MX = 1e9 + 10;
const long long int INF = 1e18;
inline long long int uceil(long long int a, long long int b) {
  return (a % b ? a / b + 1 : a / b);
}
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& v) {
  for (auto& x : v) os << x << " ";
  return os << '\n';
}
template <typename T>
ostream& operator<<(ostream& os, const set<T>& v) {
  for (auto& x : v) os << x << " ";
  return os << '\n';
}
void solve() {
  long long int n;
  cin >> n;
  long long int m = -1;
  for (long long int i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      m = i;
      break;
    }
  }
  long long int ans = 1;
  if (m != -1) ans += ((n - m) / 2);
  cout << ans << "\n";
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  solve();
  return 0;
}
