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;
template <class T>
void amax(T &x, T y) {
  if (x < y) x = y;
}
template <class T>
void amin(T &x, T y) {
  if (x > y) x = y;
}
long long int n;
const int SIZE = 1e6;
bool not_prime[SIZE];
void sieve() {
  for (long long int i = 2; i * i <= SIZE; ++i)
    if (!not_prime[i]) {
      if (n % i == 0) {
        cout << n / i << endl;
        exit(0);
      }
      for (int j = 2 * i; j <= SIZE; j += i) not_prime[j] = true;
    }
  cout << "0\n";
}
void solve() {
  cin >> n;
  sieve();
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  solve();
  return 0;
}
