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.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Scanner;

public class CodeForces
{
	public static void main(String[] args)
	{
		Scanner input = new Scanner(new BufferedReader(new InputStreamReader(System.in)));

		boolean[] arr = sieveOfEratosthenes(100001);
		ArrayList<Integer> list = new ArrayList<>();
		int[] array = new int[100001];
		array[1] = 1;
		int pos = 0;
		for (int i = 2; i < arr.length; i++)
		{
			if (arr[i])
			{
				pos++;
				array[i] = pos;
				if (i < Math.sqrt(100001))
				{
					list.add(i);
				}
			}
		}

		for (int i = 0; i < list.size(); i++)
		{
			int z = list.get(i);
			for (int j = z; j <= 100000; j += z)
			{
				if (array[j] == 0)
				{
					array[j] = i + 1;
				}
			}
		}

		int n = input.nextInt();

		StringBuilder str = new StringBuilder();
		for (int i = 2; i <= n; i++)
		{
			str.append(array[i]);
			str.append(" ");
		}
		System.out.println(str.toString());
	}

	public static boolean[] sieveOfEratosthenes(int n)
	{
		boolean prime[] = new boolean[n + 1];
		for (int i = 0; i < n; i++)
			prime[i] = true;

		for (int p = 2; p * p <= n; p++)
		{
			if (prime[p] == true)
			{
				for (int i = p * 2; i <= n; i += p)
					prime[i] = false;
			}
		}
		prime[0] = false;
		prime[1] = false;
		return prime;
	}
}