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.io.PrintWriter;
import java.util.ArrayList;
import java.util.StringTokenizer;

public class CF361B {
    public static void main(String[] args) {
        FastReader input = new FastReader();
        PrintWriter pw = new PrintWriter(System.out);

        int n = input.nextInt();
        int k = input.nextInt();

        if(n == 1 && k == 0){
            pw.println(1);
        }
        else if(k >= n){
            pw.println(-1);
        }
        else{
            int[] arr = new int[n+1];
            for(int i = 1;i <= n;i++){
                arr[i] = i;
            }

            int temp = arr[1];
            arr[1] = arr[n];
            arr[n] = temp;

            if(k == n-1){
                arr[1] = 1;
                arr[n] = n;
            }
            else if(k == 0){
                for(int i = 2;i < n;i += 2){
                    int temp2 = arr[i];
                    arr[i] = arr[i+1];
                    arr[i+1] = temp2;
                }
            }
            else if(k > 0 && k < n-1){
                if(k == n-2){

                }
                else{
                    int start = k + 2;
                    while (start + 1 <= n){
                        int temp3 = arr[start];
                        arr[start] = arr[start+1];
                        arr[start+1] = temp3;
                        start += 2;
                    }
                }
            }

            for(int i = 1;i <= n;i++){
                pw.print(arr[i] + " ");
            }
            pw.println();

        }

        pw.flush();
        pw.close();
    }
    static class FastReader
    {
        BufferedReader br;
        StringTokenizer st;

        public FastReader()
        {
            br = new BufferedReader(new
                    InputStreamReader(System.in));
        }

        String next()
        {
            while (st == null || !st.hasMoreElements())
            {
                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()
        {
            String str = "";
            try
            {
                str = br.readLine();
            }
            catch (IOException e)
            {
                e.printStackTrace();
            }
            return str;
        }
    }
}
