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;

/**
 * Created with IntelliJ IDEA.
 * User: Eugene
 * Date: 10.11.13
 * Time: 19:45
 * To change this template use File | Settings | File Templates.
 */
public class Main2 {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int k = in.nextInt();

        if (k >= n) {
            System.out.print("-1");
        } else if (n == k + 1) {
            for (int i = 1; i <= n; i++) {
                System.out.print(i + " ");
            }
        } else if (n == k + 2) {
            System.out.print("2 1 ");
            for (int i = 3; i <= n; i++) {
                System.out.print(i + " ");
            }
        } else {
            // here k < n - 1
            System.out.print(n + " ");
            for (int i = 2; i <= k + 1; i++) {
                System.out.print(i + " ");
            }
            System.out.print("1 ");
            for (int i = k + 2; i <= n - 1; i++) {
                System.out.print(i + " ");
            }
        }
    }
}
