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 Main {
    public void solve()
    {
        long kol = 0;
        int del = 1;
        String s = next();
        long n = Long.parseLong(s);
        while (true)
        {
            n = n - prost(n);
            kol++;
            if (n % 2 == 0)
            {
                del = 2;
                break;
            }
            if (n % 5 == 0)
            {
                del = 5;
                break;
            }
        }
        kol += n / 2;
        System.out.println(kol);
    }
    public static void main (String[]args){
        new Main().run();
    }

    BufferedReader br;
    StringTokenizer st;
    PrintWriter out;

    private void run () {
        try {
            br = new BufferedReader(new InputStreamReader(System.in));
            out = new PrintWriter(System.out);
            //br = new BufferedReader(new FileReader("birthday.in"));
            //out = new PrintWriter(new FileWriter("birthday.out"));
            solve();
            br.close();
            out.close();
        } catch (IOException e) {
            e.printStackTrace();
        }
    }

    private int nextInt () {
        return Integer.parseInt(next());
    }

    private String next () {
        while (st == null || !st.hasMoreTokens()) {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (IOException e) {
                e.printStackTrace();
                return "END_OF_FILE";
            }
        }
        return st.nextToken();
    }

    long prost(long n)
    {
        for (long i = 2; i <= Math.sqrt(n); i++)
        {
            if (n % i == 0) return i;
        }
        return n;
    }
}