Levko loves permutations very much. A permutation of length n is a sequence of distinct positive integers, each is at most n.

Let’s assume that value gcd(a, b) shows the greatest common divisor of numbers a and b. Levko assumes that element pi of permutation p1, p2, ... , pn is good if gcd(i, pi) > 1. Levko considers a permutation beautiful, if it has exactly k good elements. Unfortunately, he doesn’t know any beautiful permutation. Your task is to help him to find at least one of them.

Input

The single line contains two integers n and k (1 ≤ n ≤ 105, 0 ≤ k ≤ n).

Output

In a single line print either any beautiful permutation or -1, if such permutation doesn’t exist.

If there are multiple suitable permutations, you are allowed to print any of them.

Examples

Input

4 2


Output

2 4 3 1

Input

1 1


Output

-1

Note

In the first sample elements 4 and 3 are good because gcd(2, 4) = 2 > 1 and gcd(3, 3) = 3 > 1. Elements 2 and 1 are not good because gcd(1, 2) = 1 and gcd(4, 1) = 1. As there are exactly 2 good elements, the permutation is beautiful.

The second sample has no beautiful permutations.
import java.util.*;
import java.lang.*;
import java.io.*;

/* Name of the class has to be "Main" only if the class is public. */
public class Codechef
{

	public static void main (String[] args) throws java.lang.Exception
	{
		FastReader  ob=new FastReader();
        int n=ob.nextInt();
        int k=ob.nextInt();
        if(n==k)
        {
            System.out.println("-1");
            return;
        }
        int a[]=new int[n];
        if(k==n-1)
        {
            for(int i=0;i<n;i++)
            {
                a[i]=i+1;
            }
        }
        else
        {
            a[0]=n-k;
           for(int i=1;i<n-k;i++)
           {
               a[i]=i;
           }
           for(int i=n-k;i<n;i++)
           {
               a[i]=i+1;
           }
        }
     
        for(int i:a)
        System.out.print(i+" ");
	}
	
static class FastReader 
    { 
        BufferedReader br; 
        StringTokenizer st; 
  
        public FastReader() 
        { 
            br = new BufferedReader(new
                     InputStreamReader(System.in)); 
        } 
  
        String next() 
        { 
            while (st == null || !st.hasMoreElements()) 
            { 
                try
                { 
                    st = new StringTokenizer(br.readLine()); 
                } 
                catch (IOException  e) 
                { 
                    e.printStackTrace(); 
                } 
            } 
            return st.nextToken(); 
        } 
  
        int nextInt() 
        { 
            return Integer.parseInt(next()); 
        } 
  
        long nextLong() 
        { 
            return Long.parseLong(next()); 
        } 
  
        double nextDouble() 
        { 
            return Double.parseDouble(next()); 
        } 
  
        String nextLine() 
        { 
            String str = ""; 
            try
            { 
                str = br.readLine(); 
            } 
            catch (IOException e) 
            { 
                e.printStackTrace(); 
            } 
            return str; 
        } 
    } 
  
}