Problem: 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.

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.File;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Locale;
import java.util.StringTokenizer;


public class problemB {
	

	InputReader in;
	PrintWriter out;

	private void runIO() throws IOException {
//		in = new InputReader(new File("file.in"));
//		out = new PrintWriter(new File("file.out"));
		in  = new InputReader(System.in);
		out = new PrintWriter(System.out);
	}

	private void closeIO() {
		out.close();
	}

	class InputReader {
		BufferedReader bf;
		StringTokenizer st;

		InputReader(File f) throws FileNotFoundException {
			bf = new BufferedReader(new FileReader(f));
		}

		InputReader(InputStream s) {
			bf = new BufferedReader(new InputStreamReader(s));
		}

		private String next() throws IOException {
			while (st == null || !st.hasMoreElements()) {
				String s;
				try {
					s = bf.readLine();
				} catch (IOException e) {
					return null;
				}
				if (s == null)
					return null;
				st = new StringTokenizer(s);
			}
			return st.nextToken();
		}

		private int nextInt() throws NumberFormatException, IOException {
			return Integer.parseInt(next());
		}

		private long nextLong() throws NumberFormatException, IOException {
			return Long.parseLong(next());
		}

		private double nextDouble() throws NumberFormatException, IOException {
			return Double.parseDouble(next());
		}

	}

	public static void main(String[] S) throws IOException {
		// long startTime = System.currentTimeMillis();
		Locale.setDefault(Locale.US);
		new problemB().run();
		// System.out.println("Time used: "+(System.currentTimeMillis()-startTime)+" ms.");
	}

	private void run() throws IOException {
		runIO();
		solve();
		closeIO();
	}

	public int[] Permut;
	private void solve() throws IOException {
		int N = in.nextInt();
		int K = in.nextInt();
		if (K==N) {
			out.println(-1);
			return;
		}
		
		Permut = new int[N];
		for (int i=0;i<N;i++) Permut[i] = i;
		if (K%2==0) {
			if (N%2==0)
				for (int j=0;j<N-K;j+=2){
					Permut[j]++;
					Permut[j+1]--;
				}
			if (N%2!=0){
				for (int j=1;j<N-K;j+=2){
					Permut[j]++;
					Permut[j+1]--;
				}
				Permut[0] = Permut[N-K-1];
				Permut[N-K-1] = 0;
			}
		}
		if (K%2!=0) {
			if (N%2==0)
				for (int j=1;j<N-K;j+=2){
					Permut[j]++;
					Permut[j+1]--;
				}
			if (N%2!=0){
				for (int j=0;j<N-K;j+=2){
					Permut[j]++;
					Permut[j+1]--;
				}
			}
		}
		
		for (int i=0;i<N;i++)
			out.print(Permut[i]+1+" ");
		out.println();

	}
}