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 B1076 
{
    static long n;
    static int check(long num)
    {
        int c=0;
        for(int i=1;i<=(int)Math.sqrt(num);i++)
        {
            if(num%i==0)
            {
                c++;
                if(i!=num/i)
                    c++;
            }
        }
        if(c==2)
            return 1;
        else
            return 0;
    }
    public static void main(String args[])throws IOException
    {
        Scanner sc=new Scanner(System.in);
        n=sc.nextLong();
        int c=0;long min=Long.MAX_VALUE;
        for(long i=1;i<=(long)Math.sqrt(n);i++)
        {
            if(n%i==0)
            {
                if(i!=1)
                {
                    int ch=check(i);
                    if(ch==1)
                    {
                        min=Math.min(min,i);
                    }
                }
                if(i!=(n/i))
                {
                    int ch=check(n/i);
                    if(ch==1)
                    {
                        min=Math.min(min,n/i);
                    }
                }
            }
        }
        if(n%2==0)
            System.out.println(n/2);
        else
        {
            n=n-min;
            System.out.println(n/2 +1);
        }
    }
}
