Problem: 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.

Solution (in JAVA):

Solution:
import java.util.Scanner;

/**
 * Created by Andrei on 10/12/2017.
 */
public class B361 {

    public static void main(String... xxx)
    {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int k = sc.nextInt();
        int[] a = new int[n];
        if (n == k)
        {
            System.out.println(-1);
            return;
        }
        for (int i = 0; i < n; i++) {
            a[i] = i + 1;
        }
        if ((k&1) == 1)
        {
            if ((n & 1) == 1)
            {
                for (int i = 0; i < n-k;)
                {
                    int temp = a[i];
                    a[i] = a[i + 1];
                    a[i + 1] = temp;
                    i += 2;
                }
            }
            else
            {
                for (int i = 1; i < n-k;)
                {
                    int temp = a[i];
                    a[i] = a[i + 1];
                    a[i + 1] = temp;
                    i += 2;
                }
            }
        }
        else
        {
            if ((n & 1) == 1)
            {
                for (int i = 1; i < n-k;)
                {
                    int temp = a[i];
                    a[i] = a[i + 1];
                    a[i + 1] = temp;
                    i += 2;
                }
            }
            else
            {
                for (int i = 0; i < n-k;)
                {
                    int temp = a[i];
                    a[i] = a[i + 1];
                    a[i + 1] = temp;
                    i += 2;
                }
            }
        }

        StringBuilder ans = new StringBuilder();
        for (int i = 0; i < n; i++) {
            ans.append(a[i]).append(" ");
        }

        System.out.println(ans);
    }
}