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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.math.BigInteger;
import java.util.*;

import javafx.util.Pair;

public class Main
{
    //odd:
    //1 2 3 4 5 6 7 8 9  10 11
    //1 3 4 5 6 7 8 9 10 11 2

    public static void main(String[] args)
    {
        FastScanner input = new FastScanner();

        int n, k;
        n = input.nextInt();
        k = input.nextInt();
        int ans[] = new int[n + 1];
        if (n == k) {
            System.out.println("-1");
        } else if (n % 2 != 0) {

            for (int i = 1; i <= n; i++) {
                if (k + 1 >= i) {
                    ans[i] = i;
                } else if (i == n) {
                    ans[i] = k + 2;
                } else {
                    ans[i] = i + 1;
                }
            }
            if (n - 2 == k) {
                ans[1] = n;
                ans[n] = 1;
            }
            for (int i = 1; i <= n; i++) {
                System.out.print(ans[i] + " ");
            }
            System.out.println("");

        } else {
            for (int i = 1; i <= n; i++) {
                if (i % 2 == 0) {
                    ans[i] = i - 1;
                } else {
                    ans[i] = i + 1;
                }
            }

            if(k%2!=0)
            {
                ans[1]= 1;
                ans[2] =2;
                
            }
            k/=2;
            int i=3;
            while(k-->0)
            {
                ans[i]=i;
                ans[i+1]= i+1;
                i+=2;
            }
            for (int j = 1; j <=n; j++) {
                System.out.print(ans[j]+" ");
            }
            System.out.println("");
        }

    }

    static class FastScanner
    {

        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer("");

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

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

        long nextLong()
        {

            return Long.parseLong(next());
        }

        double nextDouble()
        {
            return Double.parseDouble(next());
        }

        String nextLine() throws IOException
        {
            return br.readLine();
        }
    }

}
