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.*;
import java.util.*;
import java.math.*;

public class Main
{
    public static int gcd(int a, int b)
    {
        return b==0? a:gcd(b, a%b);
    }
    public static void main(String[] args)
    {
        InputReader in = new InputReader();
        PrintWriter out = new PrintWriter(System.out);
        int n=in.nextInt();
        int k=in.nextInt();
        if(k>n-1)
            out.println("-1");
        else 
        {
            out.print(n-k);
            for(int i=1;i<n-k;i++)
                out.print(" "+i);
            for(int i=n-k+1;i<=n;i++)
                out.print(" "+i);
            out.println();
        }
        out.close();
        System.exit(0);
    }
}

class InputReader
{
    BufferedReader buf;
    StringTokenizer tok;
    InputReader()
    {
        buf = new BufferedReader(new InputStreamReader(System.in));
    }
    boolean hasNext()
    {
        while(tok == null || !tok.hasMoreElements()) 
        {
            try
            {
                tok = new StringTokenizer(buf.readLine());
            } 
            catch(Exception e) 
            {
                return false;
            }
        }
        return true;
    }
    String next()
    {
        if(hasNext()) 
            return tok.nextToken();
        return null;
    }
    int nextInt()
    {
        return Integer.parseInt(next());
    }
    long nextLong()
    {
        return Long.parseLong(next());
    }
    double nextDouble()
    {
        return Double.parseDouble(next());
    }
    BigInteger nextBigInteger()
    {
        return new BigInteger(next());
    }
    BigDecimal nextBigDecimal()
    {
        return new BigDecimal(next());
    }
}