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 May {
 
	 static int size = (int) 6e5 ;
	 static  int[] prm = new int[size] ;
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner sc = new Scanner(System.in);
		 int n = sc.nextInt();
		    int t = 0;
		    for(int i = 2; i <= n; i++){
		        if(prm[i] == 0){
		            t++;
		            for(int j = i; j <= n; j += i){
		                prm[j] = t;
		            }
		        }
		    }
		 
		    
		    for(int i = 2; i <= n; i++){
		        System.out.print(prm[i]+" ");
		    }
	}
	
private static List<Integer> prm(int n) {
		
		// TIME COMPLEXITY: O(nLogLogn)
		boolean[] primes = new boolean[n+1];
		
		for(int i=2;i<=n;++i) primes[i] = true;
		
		for(int i=2;i*i<=n;++i) {
			
			if(primes[i]) {
			for(int j=i*i;j<=n;j+=i) {
			     primes[j] = false;
			}
			}
		}
		
		List<Integer> temp = new ArrayList<>();
		for(int i=0;i< primes.length;++i)
			 if(primes[i]) prm[i] = 1;
		
		return temp;
	}
	static class Pair{
		int first;
		int second;
		Pair(int first,int second){
			this.first = first;
			this.second = second;
		}
	}
	
	
}