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;
long long md = 1000000007;
long long M = 1000000007;
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long Min(long long a, long long b) {
  if (a < b) return a;
  return b;
}
long long Max(long long a, long long b) {
  if (a > b) return a;
  return b;
}
long long power(long long a, long long b) {
  long long ans = 1;
  while (b > 0) {
    if (b & 1) ans = (ans * a) % M;
    a = (a * a) % M;
    b = b >> 1;
  }
  return ans;
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n;
  cin >> n;
  if (n & 1) {
    long long flag = 0;
    for (long long i = 3; i * i <= n; ++i) {
      if (n % i == 0) {
        flag = 1;
        n -= i;
        break;
      }
    }
    if (!flag) {
      cout << "1\n";
    } else
      cout << (n / 2) + 1 << "\n";
  } else {
    cout << n / 2 << "\n";
  }
  return 0;
}
