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 _1174C {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int[] prime = new int[n + 1];
        seive(n, prime);
        for (int i = 2; i < n + 1; i++) {
            System.out.print(prime[i] + " ");
        }
    }

    static void seive(int n, int[] prime) {
        Arrays.fill(prime, -1);
        int inc = 1;
        for (int i = 2; i * i <= n; i++) {
            if (prime[i] == -1) {
                for (int j = i; j <= n; j += i) {
                    prime[j] = inc;
                }
                inc++;
            }
        }
        for (int i = 2; i < n + 1; i++) {
            if (prime[i] == -1)
                prime[i] = inc++;
        }
    }
}