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.Scanner;

/*
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 * To change this template file, choose Tools | Templates
 * and open the template in the editor.
 */
/**
 *
 * @author KHALED
 */
public class LevkoandPermutation {
    public static void main(String[] args) {
        Scanner scan=new Scanner(System.in);
        int n=scan.nextInt();
        int k=scan.nextInt();
        if(k+1>n)
        {
            System.out.println(-1);
            return;
        }
        int[]arr=new int[n];
        arr[0]=1;
        if(n>1)
        {
            for (int i = 1; i < n; i++) 
            {
                arr[i]=i+1;                
                if(n-(i+1)>=k)
                {
                    int temp=arr[i];
                    arr[i]=arr[i-1];
                    arr[i-1]=temp;
                }
            }
        }        
        for (int b : arr) {
            System.out.print(b+" ");
        }
    }
}
