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.
/**
 * Platform     :       Codeforces
 * Url          :       https://codeforces.com/problemset/problem/361/B
 * Verdict      :       Accepted
 */

import java.util.Scanner;

public class LevkoAndPermutation{
    public static void main(String args[]){
        Scanner scan = new Scanner(System.in);
        int n = scan.nextInt();
        int k = scan.nextInt();

        int a[] = new int[n];


            for(int i = 1; i <= k; i++){
                a[n-1] = n;
                n--;
            }
            for(int i = 1; i < n; i++){
                a[i] = i;
            }
            a[0] = n;


                if(a[0] == 0){
                    System.out.println("-1");
                }
            else{
            for(int i = 0; i < n+k; i++){
                System.out.print(a[i]+" ");
            }
        }

    }
}