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.
//package practice;
import java.io.*;
import java.math.BigInteger;
import java.util.*;
import java.util.Stack;
import java.util.regex.Pattern;

public class ROUGH {
	
	public static class FastReader {
		BufferedReader br;
		StringTokenizer st;
		//it reads the data about the specified point and divide the data about it ,it is quite fast
		//than using direct 

		public FastReader() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		String next() {
			while (st == null || !st.hasMoreTokens()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (Exception r) {
					r.printStackTrace();
				}
			}
			return st.nextToken();
		}

		int nextInt() {
			return Integer.parseInt(next());//converts string to integer
		}

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

		long nextLong() {
			return Long.parseLong(next());
		}

		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (Exception r) {
				r.printStackTrace();
			}
			return str;
		}
	}
	
	public static PrintWriter out = new PrintWriter (new BufferedOutputStream(System.out));
	static long mod = (long) (1e9+7);
	static int N = (int) 1e5;
//	Scanner sc = new Scanner(System.in);
	public static void main(String[] args) {
		 FastReader sc = new FastReader();

		 long n = sc.nextLong();
		 long ans = 1 + (n-d(n))/2;
		 out.print(ans);
			 out.close();
		}

	static long d(long n)  
	{  
	   
	    if (n % 2 == 0)  
	        return 2L;  
	  
	    for (long i = 3; i * i <= n; i += 2) {  
	        if (n % i == 0)  
	            return (long)i;  
	    }  
	  
	    return n;  
	}  
	 
		
	}