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 long long mod = 1e9 + 7;
template <class S, class T>
ostream& operator<<(ostream& os, const pair<S, T> v) {
  os << "(" << v.first << ", " << v.second << ")";
  return os;
}
template <class T>
ostream& operator<<(ostream& os, const vector<T> v) {
  for (int i = 0; i < (int)v.size(); i++) {
    if (i > 0) {
      os << " ";
    }
    os << v[i];
  }
  return os;
}
template <class T>
ostream& operator<<(ostream& os, const vector<vector<T>> v) {
  for (int i = 0; i < (int)v.size(); i++) {
    if (i > 0) {
      os << endl;
    }
    os << v[i];
  }
  return os;
}
map<long long, long long> mp;
void prime_factorization(long long x) {
  bool flag = true;
  while (flag) {
    long long sqx = sqrt(x) + 1;
    for (long long j = 2; j < sqx + 1; j++) {
      if (x % j == 0) {
        mp[j]++;
        x /= j;
        break;
      }
      if (j == sqx) flag = false;
    }
  }
  if (x != 1) {
    mp[x]++;
  }
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  long long n;
  cin >> n;
  if (n % 2 == 0) {
    cout << n / 2 << endl;
  } else {
    prime_factorization(n);
    long long val = mp.begin()->first;
    n -= val;
    cout << n / 2 + 1 << endl;
  }
  return 0;
}
