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.*;
import java.lang.*;
import java.io.*;
public class Test
{
	
	
	
	
	public static void main (String[] args)
	{
		
			Scanner sc=new Scanner(System.in);
		
			int j,n;
			n=sc.nextInt();
			
			int a[]=new int[n];
			//a[0]=0;
			
			Arrays.fill(a,1);
			
			
			
			n+=1;
			int al=1;
			
			int prime[] = new int[n+1]; 
	        
			for(int i=0;i<n;i++) 
	            prime[i] = 1; 
	          
	        for(int p = 2; p<=n; p++) 
	        { 
	            // If prime[p] is not changed, then it is a prime 
	            if(prime[p] == 1) 
	            { 
	            	//a[p-2]=al;
	                // Update all multiples of p 
	                for(int i = p; i <= n; i += p)
	                {
	                    prime[i] = 0;
	                    a[i-2]=al;
	                }
	                al++;
	            }
	        } 
	          
	     
	        
	       for(int i=0;i<n-2;i++)
	        	System.out.print(a[i]+" ");
	   
		
	}
}