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.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Scanner;


public class EhabColor {

  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    PrintWriter out = new PrintWriter(System.out);
    int n = sc.nextInt();
    ArrayList<Integer> primes = sieve(n);
    int[] arr = new int[n+1];
    int cnt = 1;
    for(Integer p : primes) {
      for(int i = p; i <= n; i+=p) {
        if(arr[i]==0) arr[i] = cnt;
      }
      cnt++;
    }
    for(int i = 2; i <= n; i++) {
      out.print(arr[i] + " ");
    }
    out.println();
    out.close();
  }
  
  static ArrayList<Integer> sieve(int n) {
    boolean[] prime = new boolean[n+1];
    Arrays.fill(prime,true);
    for(int p=2; p*p <= n; p++) {
      if(prime[p]) {
        for(int i = p*2; i <= n; i+=p) {
          prime[i] = false;
        }
      }
    }
    
    ArrayList<Integer> primes = new ArrayList();
    for(int i = 2; i <= n; i++) {
      if(prime[i]) primes.add(i);
    }
    return primes;
  }

}