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;

public class p2_210 {
    private static int GCD(int number1, int number2) {
        if (number2 == 0) {
            return number1;
        }
        return GCD(number2, number1 % number2);
    }

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
//      while (true) {
            int n = in.nextInt();
            int k = in.nextInt();
            if (k == n)
                System.out.println(-1);
            else {
                int[] list = new int[n];
                for (int i = n - 1; i >= (n - k); i--)
                    list[i] = i + 1;
                int start = n - k;
                if (start % 2 == 0) {
                    for (int i = 0; i < start; i++) {
                        if ((i + 1) % 2 == 0)
                            list[i-1] = i + 1;
                        else
                            list[i + 1] = i + 1;
                    }
                } else {
                    list[0] = 1;
                    for (int i = 1; i < start; i++) {
                        if ((i + 1) % 2 == 0)
                            list[i + 1] = i + 1;
                        else
                            list[i-1] = i + 1;
                    }
                }
            for (int i = 0; i < list.length - 1; i++) {
                System.out.print(list[i] + " ");
            }
            System.out.println(list[list.length - 1]);
//              int count = 0;
//              for (int i = 0; i < list.length; i++) {
//                  if (GCD(i + 1, list[i]) > 1)
//                      count++;
//              }
//              System.out.println(count + "  " + k);
//          }
        }
    }
}
