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 Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		while (sc.hasNext()) {
			int n = sc.nextInt();
			int arr[] = new int[n + 1];
			for (int i = 2; i <= n; i++) {
			}
			int sum = 1;
			arr[2] = 1;
			for (int i = 3; i <= n; i++) {
				boolean flag = false;
				for(int j=2;j<=Math.sqrt(i);j++) {
					if(i%j==0) {
						arr[i]=arr[j];
						flag = true;
						break;
					}
				}
				if(!flag) {
					arr[i] = ++sum;
				}
			}
			for (int i = 2; i <= n; i++) {
				if(i!=n) {
					System.out.print(arr[i]+" ");
				}else {
					System.out.print(arr[i]);
				}
			}
			System.out.println();
		}
	}
}