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.*;

public class Solve{
     static boolean[] prime=new boolean[100005];
   
	public static void main(String[] args) {
		Scanner sc=new Scanner(System.in);
	
         int n=sc.nextInt();

		StringBuilder sb=new StringBuilder();
         if(n==2){System.out.println("1");}
         else if(n==3){System.out.println("1 2");}
         else if(n==4){System.out.println("1 2 1");}
         else{ int a=2;
                 int[] ar=new int[n+1];
               ar[2]=1;
               ar[3]=2;
               ar[4]=1;
               seive();
             
               for(int i=5;i<=n;i++){
                   if(prime[i]==true){
                       a++;
                       ar[i]=a;
                   }
                   else{
                       ar[i]=ar[odd(i)];
                   }
                  
               }
                for(int i=2;i<=n;i++){
                  sb.append(ar[i]+" ");
                  
               }
               System.out.println(sb);
         }
	}

	static int odd(int n){
	    int ans=1;
	    for(int i=2;i*i<=n;i++){
	        if(n%i==0){
	            if(prime[i]==true){
	                ans=Math.max(ans,i);
	            }
	             if(prime[n/i]==true){
	                ans=Math.max(ans,n/i);
	            }
	        }
	    }
	    return ans;
	}
	
	static void seive(){
	    int n=prime.length-1;
	     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; 
         
	}
	
        
	}
	}
	
}