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.
import java.io.*;
import java.util.*;
 
public class codeforce {
 
	static long m = (long) (1e9 + 7);
 
	public static void main(String[] args) throws IOException {
		Scanner scn = new Scanner(System.in);
		StringBuilder sb = new StringBuilder();
		long n = scn.nextLong();
		if (n % 2 == 0)
			sb.append(n / 2);
		else {
			ArrayList<Long> l = primeFactors(n);
			long ans = 1;
			n -= l.get(0);
			ans += n / 2;
			sb.append(ans);
		}
		sb.append("\n");
		System.out.print(sb);
	}
 
	public static ArrayList<Long> primeFactors(long n) {
		ArrayList<Long> al = new ArrayList<>();
		while (n % 2 == 0) {
			al.add(2L);
			n /= 2;
		}
		for (long i = 3; i <= Math.sqrt(n); i += 2) {
			while (n % i == 0) {
				al.add(i);
				n /= i;
			}
		}
		if (n > 2)
			al.add(n);
		return al;
	}
 
	static class Scanner {
 
		StringTokenizer st;
		BufferedReader br;
 
		public Scanner(InputStream s) {
			br = new BufferedReader(new InputStreamReader(s));
		}
 
		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}
 
		public int nextInt() throws IOException {
			return Integer.parseInt(next());
		}
 
		public long nextLong() throws IOException {
			return Long.parseLong(next());
		}
	}
}