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.io.*;

public class Main {
    public static void main(String[] args) {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        OutputStream output = System.out;
        PrintWriter pr = new PrintWriter(output);
        try {
            String []line = br.readLine().split("\\s+");
            int n = Integer.parseInt(line[0]);
            int k = Integer.parseInt(line[1]);

            if (n == k) {
                pr.println(-1);
            }
            else {
                pr.print((n - k) + " ");
                pr.flush();
                for (int i = 1 ; i < n - k ; i ++) {
                    pr.print(i + " ");
                    pr.flush();
                }
                for (int i = n - k + 1 ; i <= n ; i ++) {
                    pr.print(i + " ");
                    pr.flush();
                }
            }
            pr.close();
        } catch (IOException e) {
            e.printStackTrace();
        }
    }
}
