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.InputStreamReader;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
	public static void main(String[] args) {
		InputStream inputStream = System.in;
		OutputStream outputStream = System.out;
		InputReader in = new InputReader(inputStream);
		PrintWriter out = new PrintWriter(outputStream);
		TaskB solver = new TaskB();
		solver.solve(1, in, out);
		out.close();
	}
}

class TaskB {
    public void solve(int testNumber, InputReader in, PrintWriter out) {
        int n=in.nextInt();
        int k=in.nextInt();
        if(k>=n) {
            out.println(-1);
            return;
        }

        int ar[]=new int[n];
        for(int i=0;i<n;i++) {
            ar[i]=i+1;
        }
        int bad=n-k;
        bad--;
        int i=1;
        while(bad>=2) {
            int t=ar[i];
            ar[i]=ar[i+1];
            ar[i+1]=t;
            bad-=2;
            i+=2;
        }
        if(bad==1) {
            int t=ar[0];
            ar[0]=ar[n-1];
            ar[n-1]=t;
        }

        for(int j=0;j<n;j++) {
            if(j!=0)
                out.print(" ");
            out.print(ar[j]);
        }
        out.printf("\n");
    }
}

class InputReader {
    StringTokenizer st;
    BufferedReader in;
    public InputReader(InputStream ins)
    {
        in = new BufferedReader(new InputStreamReader(ins));
    }

    public String nextToken()
    {
        while(st==null || !st.hasMoreTokens())
        {
            try {
                st=new StringTokenizer(in.readLine());
            } catch (IOException e) {
                e.printStackTrace();  //To change body of catch statement use File | Settings | File Templates.
            }
        }
        return st.nextToken();
    }
    public int nextInt()
    {

        return Integer.parseInt(nextToken());
    }

    }

