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.Arrays;
import java.util.Scanner;

/**
 * Created by  on 04/06/19.
 */
public class HelloWorld6 {
    public static void main(String[] args) {
        Scanner s = new Scanner(System.in);
        int n = s.nextInt();
//        int[] arr = new int[n - 1];
        int ele = 0;

        boolean[] isPrime = sieve(n);
        int[] arr = new int[n + 1];
        for (int i = 2; i <= n; i++) {
            if(!isPrime[i]) {
                continue;
            } else {
                ele++;
                for (int j = 1; j * i <= n; j++) {
                    arr[i * j] = ele;
                }
            }
        }
        StringBuilder ans = new StringBuilder();
        for (int i = 2; i <= n; i++) {
            ans.append(arr[i]).append(" ");
        }
        System.out.println(ans.toString());
    }
    private static boolean[] sieve(int n) {
        boolean[] isPrime = new boolean[n + 1];
        Arrays.fill(isPrime, true);
        for (int i = 2; i <= n; i++) {
            if(isPrime[i]) {
                for (int j = 2; j * i <= n; j++) {
                    isPrime[i*j] = false;
                }
            }
        }

        return isPrime;
    }
}