Problem: 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.

Solution (in JAVA):

Solution:
import java.util.Scanner;


public class vPerm {

    public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        
        int n=sc.nextInt();
        int m=sc.nextInt();
        
        if(m==n)
            System.out.println(-1);
        else if(m==n-1){
            for(int i=1;i<=n;i++){
                System.out.print(" "+i);
            }
            System.out.println();
        }else{
            if(n%2==0){
                if(m%2==0){
                    System.out.print(" 2 1");
                    for(int i=3;i<=m+2;i++){
                        System.out.print(" "+i);
                    }
                    for(int i=m+3;i<=n;i+=2){
                        System.out.print(" "+(i+1)+" "+(i));
                    }
                    System.out.println();
                }else{
                    for(int i=1;i<=m+1;i++){
                        System.out.print(" "+i);
                    }
                    for(int i=m+2;i<=n;i+=2){
                        System.out.print(" "+(i+1)+" "+i);
                    }
                }
                System.out.println();
            }else{
                if(m%2==0){
                    for(int i=1;i<=m+1;i++){
                        System.out.print(" "+i);
                    }
                    for(int i=m+3;i<=n;i+=2){
                        System.out.print(" "+(i)+" "+(i-1));
                    }
                    System.out.println();
                }else{
                    System.out.print(" "+n);
                    for(int i=2;i<=m+1;i++){
                        System.out.print(" "+i);
                    }
                    System.out.print(" "+1);
                    for(int i=m+3;i<n;i+=2){
                        System.out.print(" "+(i-1)+" "+i);
                    }
                }
                System.out.println();
            }
//          if(m%2==1){
//              for(int i=1;i<=m+1;i++){
//                  System.out.print(" "+i);
//              }
//              
//              for(int i=m+2;i<=n;i+=2){
//                  System.out.print(" " +(i+1)+" "+i);
//              }
//          
//              System.out.println();
//          }else{
//              System.out.print(" "+n);
//              for(int i=2;i<=m+1;i++){
//                  System.out.print(" "+i);
//              }
//              System.out.print(" "+1);
//              for(int i=m+2;i<n;i+=2){
//                  System.out.print(" " +(i+1)+" "+i);
//              }
//              System.out.println();
//          }
        }
    }

}