Problem: You're given an integer n. For every integer i from 2 to n, assign a positive integer a_i such that the following conditions hold:

  * For any pair of integers (i,j), if i and j are coprime, a_i ≠ a_j. 
  * The maximal value of all a_i should be minimized (that is, as small as possible). 



A pair of integers is called [coprime](https://en.wikipedia.org/wiki/Coprime_integers) if their [greatest common divisor](https://en.wikipedia.org/wiki/Greatest_common_divisor) is 1.

Input

The only line contains the integer n (2 ≤ n ≤ 10^5).

Output

Print n-1 integers, a_2, a_3, …, a_n (1 ≤ a_i ≤ n). 

If there are multiple solutions, print any of them.

Examples

Input


4


Output


1 2 1 

Input


3


Output


2 1

Note

In the first example, notice that 3 and 4 are coprime, so a_3 ≠ a_4. Also, notice that a=[1,2,3] satisfies the first condition, but it's not a correct answer because its maximal value is 3.

Solution (in JAVA):

Solution:
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Scanner;
public class ehab
{
	public static void main (String[] args)
	 {
		Scanner sc=new Scanner(System.in); 
		int t,i;
		t=1;
		while(t>0)
		{
			t--;
			
			int n=sc.nextInt();
			ArrayList<Integer> pr=new ArrayList();
			for(i=2;i<=n;i++)
			{
				if(isp(i))
					pr.add(i);
			}
			int c[]=new int[n+1];
			int k=1;
			for(i=0;i<pr.size();i++)
				c[pr.get(i)]=k++;
			for(i=2;i<=n;i++)
			{
				if(c[i]==0)
				{
					for(int j=2;j*j<=i;j++)
					{
						if(i%j==0)
						{
							System.out.print(c[j]+" ");
							break;
						}
					}
				}
				else
					System.out.print(c[i]+" ");
			}	
			
		}
	} 
	static boolean isp(int n)
	{
		
		int i;
		for(i=2;i*i<=n;i++)
		{
			if(n%i==0)
				return false;
		}
		return true;
	}
}