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 c{


public static int factor(int n){

for(int i=2;i<=Math.sqrt(n);i++){
	if(n%i==0) return i;

}
 return 0;
}
static boolean prime[] = new boolean[100001]; 
 public static void s() 
    { 
        // Create a boolean array "prime[0..n]" and initialize 
        // all entries it as true. A value in prime[i] will 
        // finally be false if i is Not a prime, else true. 
        
        for(int i=0;i<=100000;i++) 
            prime[i] = true; 
          
        for(int p = 2; p*p <=100000; p++) 
        { 
            // If prime[p] is not changed, then it is a prime 
            if(prime[p] == true) 
            { 
                // Update all multiples of p 
                for(int i = p*p; i <= 100000; i += p) 
                    prime[i] = false; 
            } 
        } 
          }
        // Print all prime numbers 
       

public static void main(String args[]){
Scanner sc = new Scanner(System.in);
int k=0;
 int flag = 0;
int n=sc.nextInt();
s();  

int[] f=new int[n-1];
f[0]=1;

int cnt=2;
for(int i=3;i<n+1;i++){
if(prime[i]){
	f[i-2]=cnt;
	cnt++;
}
else if(i%2==0){f[i-2]=1;
}
else{
f[i-2]=f[factor(i)-2];
}

}
for(int i=0;i<n-1;i++)
System.out.print(f[i]+" ");

}


}