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.io.*;
import java.util.Scanner;
import java.util.*;
public class  p2
 {
   public static void main(String args[])
   {
    int n,k;
	 Scanner in = new Scanner(System.in);
	  n=in.nextInt();
	  k=in.nextInt();
	  int alter=n-k;
	  if(k>n-1) System.out.println(-1);
	   else if(alter==1 && n==2)
	    {
		  System.out.println(1+" "+2);
		}
		
		
		else
		 {
		  if(alter==1)alter--;
		 int arr[]=new int[n+10];
		  
		  int i=1;
		   if(alter%2==1 && alter!=1)
		    {
			   int temp1=1;
			   int temp2=2;
			   int temp3=3;
			   arr[1]=2;
			   arr[2]=3;
			   arr[3]=1;
			   alter-=3;
			   i=4;
			     
			}
		  while(i!=n+1)
		   {
		  // System.out.println("here " +i);
		    if(alter!=0)
			 {
			 //System.out.println("swapin ");
			   alter-=2;
			    arr[i]=i+1;
				arr[i+1]=i;
			   i+=2;
			 }
			 else 
			 {
			   arr[i]=i;
			    i++;
			 }
		   }
		   
		     for(i=1;i<=n;i++)
			  System.out.print(arr[i]+" ");
			  
		   
		 }
		 }
 }