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.Scanner;

public class Problemc {

	public static void main(String[] args) {
		// TODO Auto-generated method stub

		Scanner sc = new Scanner(System.in);

		int a = sc.nextInt();
		int[] arr = new int[a + 1];

		int k = 0;
		for (int i = 2; i < arr.length; i++) {
			if (arr[i] != 0) {
				continue;
			}
			if (isprime(i)) {
				k++;
				int j = i;
				while (j < arr.length) {
					arr[j] = k;
					j = j+i;
				}
			}
		}
		for(int i=2;i<arr.length;i++) {
			System.out.println(arr[i]);
		}

	}


	private static boolean isprime(long prime) {

		// TODO Auto-generated method stub
		if (prime == 2) {
			return true;
		}

		if (prime <= 1) {
			return false;
		} else if (prime % 2 == 0) {
			return false;
		} else {

			double range = Math.sqrt(prime) + 1;

			for (int i = 3; i <= range; i++) {

				if (prime % i == 0) {
					return false;
				}

			}

		}

		return true;
	}

}