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.Hashtable;
import java.util.Scanner;

public class CF1174C {

	public static void main(String[] args) {
		Scanner s = new Scanner(System.in);
		int n = s.nextInt();
		int[] primeGroup = new int[n-1];
		Hashtable<Integer, Integer> groupMap = new Hashtable<>();
		int groupIndex = 1;
		for (int i = 2; i <= n; i++) {
			int group=primeFactorization(i);
			if(!groupMap.containsKey(group)) {
				groupMap.put(group, groupIndex);
				groupIndex++;
			}
			primeGroup[i-2]=group;
		}
		for (int i = 0; i < primeGroup.length; i++) {
			System.out.print(groupMap.get(primeGroup[i])+" ");
		}
	}
	public static int primeFactorization(int x) {
		int i=3;
		int c;
		c=x;
		while((c%2)==0) {
			return 2;
		}
		while(i<=(Math.sqrt(c)+1)) {
			if((c%i)==0) {
				return i;
			}else {
				i+=2;
			}
		}
		return x;
	}
}