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 = 3e5 + 1000;
const long long INFINF = 900372036851194844;
const int INF = 1e9 + 7;
template <class T>
T gcd(T a, T b) {
  return b ? gcd(b, a % b) : a;
}
void prtf() {}
template <typename Arg1, typename... Args>
void prtf(const Arg1& arg1, const Args&... args) {
  cout << arg1;
  prtf(args...);
  exit(0);
}
int main() {
  long long n;
  long long ct = 0;
  cin >> n;
  long long d = 0, q = sqrt(n);
  for (int i = 2; i <= q; ++i) {
    for (int j = 2; j * j <= i; ++j)
      if (i % j == 0 && i > 2) continue;
    if (n % i == 0) {
      d = i;
      break;
    }
  }
  if (d == 0)
    ct = n;
  else
    ct = n / d;
  prtf(ct);
  return 0;
}
