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.util.*;
import java.io.*;

public class divsub {
	static BufferedReader br;
	static StringTokenizer tokenizer;

	public static void main(String[] args) throws Exception {
		br = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter pw = new PrintWriter(System.out);
		long n = nextLong(); long div = n;
		int max = (int) Math.ceil(Math.sqrt(n));
        for (int i = 2; i <= max; i ++) {
            if ((n / i * i) == n) {
                div = i;
                break;
            }
        }
        long add = 0;
        if((n & 1) == 1) {
            n -= div;
            add = 1;
        }
        pw.println((add + (n / 2)));
        pw.close();
	}

	public static String next() throws IOException {
		while (tokenizer == null || !tokenizer.hasMoreTokens()) {
			String line = br.readLine();
			if (line == null)
				throw new IOException();
			tokenizer = new StringTokenizer(line);
		}
		return tokenizer.nextToken();
	}

	public static int nextInt() throws IOException {
		return Integer.parseInt(next());
	}

	public static long nextLong() throws IOException {
		return Long.parseLong(next());
	}

	public static double nextDouble() throws IOException {
		return Double.parseDouble(next());
	}
}
