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 B
{

	public static void main(String[] args) throws Exception
	{
		Scanner sc = new Scanner(System.in);
		PrintWriter out  = new PrintWriter(System.out);
		
		long x = sc.nextLong();
		
		if(x % 2 == 0 )
			out.println(x / 2);
		else 
		{
			long min = x;
			for(long i = 2 ; i * i <= x ; i++)
				if(x % i == 0)
					min = Math.min(min, Math.min(x / i, i)) ;
			
			out.println(1 + (x - min) / 2);
		}
		
		out.flush();
		out.close();
		
		

	}
	static class Scanner 
	{
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream s){	br = new BufferedReader(new InputStreamReader(s));}
		
		public Scanner(String path) throws Exception{	br = new BufferedReader(new FileReader(path));}

		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());}


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

	}

	static void shuffle(int[] a)
	{
		int n = a.length;
		for(int i = 0; i < n; i++)
		{
			int r = i + (int)(Math.random() * (n - i));
			int tmp = a[i];
			a[i] = a[r];
			a[r] = tmp;
		}
	}



}
