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:
//package pack1;
import java.util.*;
public class Codechef1 {
	
	public static void main(String[] args) {
		// TODO Auto-generated method stub
        Scanner s=new Scanner(System.in);
        int n=s.nextInt();
        TreeMap<Integer,Integer> hs=new TreeMap<Integer,Integer>();
        int c=0,q=0;
         for(int i=2;i<=n;i=i+2)
         {
        	 q++;
        	 int p=i;
        	 int j=2;
        	 if(!hs.containsKey(p))
        	 {c++;
        	// System.out.println(i+" "+c);
        	 while(p<=n)
        	 {
        		 //if(!hs.containsKey(p))
        		 hs.put(p,c);
        		 p=i*j;
        		 j++;
        	 }
        	 }
        	 if(i==2)
            	 i=i-1;
        	 
         }

         for(int i=2;i<=n;i++)
         {
        	 if(hs.containsKey(i))
        		 System.out.print(hs.get(i)+" ");
        	 else
        		 {c++;
        		 System.out.print(c+" ");}
         }
         System.out.println();
        //  System.out.println(hs);
        
        	
	}

}