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 gcd(long long a, long long b) { return !b ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return (a / gcd(a, b)) * b; }
enum dir {
  DOWN = 0,
  UP,
  RIGHT,
  LEFT,
  DOWN_RIGHT,
  DOWN_LEFT,
  UP_RIGHT,
  UP_LEFT
};
int dir_r[] = {1, -1, 0, 0, 1, 1, -1, -1};
int dir_c[] = {0, 0, 1, -1, 1, -1, 1, -1};
bool p(long long s) {
  if (s == 1 || (s % 2 == 0 && s != 2)) return 0;
  for (int i = 2; i <= sqrt(s); i++) {
    if (s % i == 0) return 0;
  }
  return 1;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  if (p(n)) return cout << 1, 0;
  long long ans = 1e12;
  for (long long i = 1; i <= sqrt(n); i++) {
    if (n % i == 0) {
      if (p(i)) {
        ans = min(ans, i);
      }
      if (p(n / i)) ans = min(ans, i);
    }
  }
  if (ans == 1e12)
    cout << 0;
  else {
    n -= ans;
    cout << n / 2 + 1;
  }
  return 0;
}
