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.io.*;
public class primes {
	static ArrayList<Integer> arr;
	public static void main(String[] args) throws Exception {
		//Scanner in = new Scanner(new File("primes.in"));
		Scanner in = new Scanner(System.in);
		arr = new ArrayList<Integer>();
		int n = in.nextInt();
		for (int i = n; i >1; i--) {
			if(isPrime(i))
				arr.add(i);
		}
		
		int[] x = new int[n+1];
		for (int i = 0; i < arr.size(); i++) 
		{
			x[arr.get(i)] = arr.size()-i;
		}
		for(int i=2;i<n+1;i++)
		{
			if(x[i] == 0)
			{
				x[i] = x[lowestPrimeFactor(i)];
			}
		}
		
		
		for(int i=2;i<n+1;i++)
			System.out.print(x[i] + " ");

		
		
	}
	public static boolean isPrime(int n)
	{
		int max = (int) Math.sqrt(n) +1;
        for (int i = 2; i < max; i++) 
        {
        	if(n%i == 0)
        	return false;        
        }
        return true;
	}
	public static int lowestPrimeFactor(int n)
	{
		int max = (int) Math.sqrt(n) +1;
		for(int i=2;i<max;i++)
		{
			if(n%i == 0)
				return i;
		}
		return n;
	}
}