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 MyClass {
    public static void main(String args[]) {
     Scanner in=new Scanner(System.in);
     int n=in.nextInt();
     boolean prime[]=new boolean[n+1];
     Arrays.fill(prime,true);
     int ans[]=new int[n+1];
     for(int i=2;i<=n;i++) ans[i]=i-1;
     for(long i=2;i<=n;i++){
         if(prime[(int)i]){
             for(long j=i*i;j<=n;j+=i) { prime[(int)j]=false; ans[(int)j]=ans[(int)i]; }
         }
     }
     //System.out.println(Arrays.toString(ans));
     int temp[]=new int[n-1];
     int c=0;
     Arrays.fill(temp,Integer.MAX_VALUE);
     HashSet<Integer> h=new HashSet<>();
     for(int i=0;i<n-1;i++){
         if(h.contains(ans[i+2])) continue;
         temp[c++]=ans[i+2];
         h.add(ans[i+2]);
     }
     int index[]=new int[n+1];
     Arrays.sort(temp);
     for(int i=0;i<n-1;i++){
         if(temp[i]==Integer.MAX_VALUE) break;
         index[temp[i]]=i+1;
     }
     for(int i=2;i<=n;i++){
         System.out.print(index[ans[i]]+" ");
     }
     System.out.println("");
    }
}