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.Scanner;

public class Main {
	
	public static void main(String[] args)
	{
		Scanner s=new Scanner(System.in);
		
		int n=s.nextInt();
		
		boolean[] ans=new boolean[n+1];
		
		sieve(n,ans);
		
		int[] arr=new int[n+1];
		
		int count=1;
		
		for(int j=2;j<=n;j++)
		{
			if(ans[j])
			{
				int val=j;
				
				while(val<=n)
				{
					if(arr[val]==0)
					{
					arr[val]=count;
					}
					val=j+val;
				}
				
				count++;
			}
		}
		
		for(int j=2;j<=n;j++)
		{
			System.out.print(arr[j]+" ");
		}
		
	}
	
	public static void sieve(int n,boolean[] prime) 
    {  
        for(int i=0;i<=n;i++) 
            prime[i] = true; 
          
        for(int p = 2; p*p <=n; p++) 
        { 
            if(prime[p] == true) 
            { 
                for(int i = p*p; i <= n; i += p) 
                    prime[i] = false; 
            } 
        } 
    } 
	
}