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 CP {


    static int[] arr = new int[100005];
    static int[] var = new int[100005];
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        for(int i=1;i<=n;i++) {
            arr[i] = var[i] = 0;
        }
        sieve(n);
        for(int i = 2; i<=n; i++) {
            System.out.print(arr[i] + " ");
        }
    }

    static void sieve(int n) {
        int cnt = 0;
        for(int i = 2; i*i <= n; i++) {
            if(arr[i] == 0) {
                cnt++;
                arr[i] = cnt;
                var[i] = 1;
                for(int j = i*i; j<=n; j+=i) {
                    if(arr[j] == 0){
                        arr[j] = cnt;
                    }
                }
            }
        }
        for(int i=2; i<=n; i++) {
            if(arr[i]==0) {
                cnt++;
                arr[i] = cnt;
                var[i] = 1;
            }
            var[i] += var[i-1];
        }
    }
}