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 <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << " : " << arg1 << std::endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
const int L = 1e7 + 7;
map<long long, long long> counter;
int lps[L];
void pre() {
  lps[1] = 1;
  for (int i = 2; i < 10000001; ++i) {
    if (lps[i]) continue;
    for (int j = i; j <= 10000000; j += i) {
      lps[j] = i;
    }
  }
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, ans = 0;
  cin >> n;
  int tmp = sqrt(n);
  if (n % 2 == 0) {
    cout << n / 2;
    return 0;
  }
  for (long long i = 3; i <= tmp + 1; i += 2) {
    if (n % i == 0) {
      cout << (n - i) / 2 + 1;
      return 0;
    }
  }
  cout << "1";
  return 0;
}
