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.
/* 
    Author: Aman Patel
    Date: 28-05-2021
*/


import java.io.*;
import java.util.StringTokenizer;

public class LevkoAndPermuatation {

    public static void main(String[] args) {

        FastScanner fs = new FastScanner();
        PrintWriter out = new PrintWriter(System.out);

        int n, k;
        n = fs.nextInt();
        k = fs.nextInt();

        if (k >= n)
            out.println(-1);
        else if (k == n - 1) {
            for (int i = 1; i <= n; i++)
                out.print(i + " ");
        } else {

            int[] result = new int[n + 5];
            result[1] = n;
            int i;
            for (i = 2; i <= (1 + k); i++)
                result[i] = i;

            int len = n - k - 1;
            if ((len & 1) == 0) {
                for (i = k + 2; i <= n; i += 2) {
                    result[i] = i + 1;
                    result[i + 1] = i;
                }
                result[n - 1] = 1;
            } else {
                for (i = k + 2; i < n; i += 2) {
                    result[i] = i + 1;
                    result[i + 1] = i;
                }
                result[n] = 1;
            }

            for (i = 1; i <= n; i++)
                out.print(result[i] + " ");
        }

        out.close();
    }

    static class FastScanner {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer("");

        public String next() {
            while (!st.hasMoreElements())
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            return st.nextToken();
        }

        int[] readArray(int n) {
            int[] a = new int[n];
            for (int i = 0; i < n; i++) a[i] = nextInt();
            return a;
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        Long nextLong() {
            return Long.parseLong(next());
        }
    }

    public static PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
}
