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.ArrayList;
import java.util.Collections;
import java.util.Scanner;

public class gcd {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        String line[] = in.nextLine().split(" ");
        int n = Integer.parseInt(line[0]);
        int k = Integer.parseInt(line[1]);
        if (n == k)
            System.out.println(-1);
        else {
            ArrayList<Integer> w= new ArrayList<Integer>();
            int s=n;
            for (int i = 0; i < k; i++) {
                w.add(n--);
            }
            Collections.reverse(w);
            ArrayList<Integer> r= new ArrayList<Integer>();

            for (int i = 2; i < n+1 ; ++i) {
                r.add(i);
            }
            r.add(1);
            r.addAll(w);
            for( int i =0;i<r.size()-1;i++)
            System.out.print(r.get(i)+ " " );
            System.out.println(r.get(r.size()-1));
        }

    }
}
