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 CodeForce {
	static void sieve(int n)
	{
		boolean flags[] = new boolean[n+1];
		int code[] = new int[n+1];
		int p=2;
		while(p <= Math.sqrt(n)) {
			cross(flags,p);
			p = getNext(flags,p);
		}
		int x = 1;
		for(int i=2;i<flags.length;i++)
			if(!flags[i]) {

				for(int j=i;j<code.length;j+=i)
					if(code[j] == 0)
						code[j] = x;
				x+=1;
			}
		for(int i=2;i<code.length;i++) {
			System.out.print(code[i]+" ");
		}
	}
	static void cross(boolean flags[],int p)
	{
		for(int i=p*p;i<flags.length;i+=p)
			flags[i]= true;
	}
	static int getNext(boolean flags[],int p)
	{
		int next = p+1;
		while (next < flags.length && flags[next])
			next++;
		return next;
	}

	public static void main(String[] args) {
		Scanner scan = new Scanner(System.in);
		ArrayList<Integer> primes = new ArrayList<>();
		int n = scan.nextInt();
		sieve(n);
	}

}