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;
bool c[100010];
long long p[100001];
string s;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n;
  cin >> n;
  int index = 0;
  for (long long i = 2; i <= 100000; i++) {
    if (!c[i]) {
      p[index++] = i;
      for (long long j = i * i; j <= 100000; j += i) {
        c[j] = true;
      }
    }
  }
  long long count = 0;
  while (true) {
    bool flag = false;
    for (int i = 0; i < index; i++) {
      if (n <= p[i]) break;
      if (n % p[i] == 0) {
        flag = true;
        n -= p[i];
        break;
      }
    }
    count++;
    if (!flag) {
      cout << count;
      return 0;
    }
    if (n % 2 == 0) {
      cout << n / 2 + count;
      return 0;
    }
  }
}
