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 kotlin.reflect.jvm.internal.impl.resolve.constants.KClassValue;

import java.util.Arrays;
import java.util.Scanner;

public class Sum {

    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);

       int n,k;
       n = input.nextInt();
       k=input.nextInt();
       int A[]=new int[n+2];
       if(n-1<k) System.out.println("-1");
       else{
           for(int i=1; i<=k+1; i++){
               ///System.out.print(i+" ");
               A[i]=i;
           }
           for(int i=k+2; i<n; i+=2){
               A[i]=i+1;
               A[i+1]=i;
           }
           if((n-k)%2==0){
               A[1]=n;
               A[n]=1;
           }
           for(int i=1; i<=n;i++){
               System.out.print(A[i]+" ");
           }
           System.out.println();
       }



    }
}