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 whatever; // don't place package name! */

import java.util.*;
import java.lang.*;
import java.io.*;

/* Name of the class has to be "Main" only if the class is public. */
public class Main
{
	public static void main (String[] args) throws java.lang.Exception
	{
				Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int prime[] = new int[1000001];
		int arr[] = new int[n+1];
		int count = 0;
		for(int i=2;i*i<=1000000;i++){
			if(prime[i]==0){
			for(int j=2*i;j<=1000000;j+=i){
				prime[j]=1;
			}
 
			}
		}
		for(int i=2;i<=n;i++){
			if(prime[i]==0){
				count++;
				for(int j=i;j<=n;j+=i){
					if(arr[j]==0)
					arr[j]=count;
 
				}
			}
 
		}
		for(int i=2;i<=n;i++)
		System.out.print(arr[i]+" ");
	}
}