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 static java.lang.Math.*;
import static java.util.Arrays.*;
import java.util.*;
import java.io.*;

public class Main {

	void solve() {
		int n = sc.nextInt();
		int k = sc.nextInt();

		if (k >= n) {
			out.println(-1);
			return;
		}

		int[] a = new int[n];
		for (int i = 0; i < n; i++) {
			a[i] = i + 1;
		}

		rot(a, 0, n - k - 1);
		print(a);
	}

	void rot(int[] a, int i, int j) {
		int t = a[i];
		for (int k = i; k < j; k++) {
			a[k] = a[k+1];
		}
		a[j] = t;
	}

	void print(int[] a) {
		out.print(a[0]);
		for (int i = 1; i < a.length; i++) out.print(" " + a[i]);
		out.println();
	}

	static void tr(Object... os) {
		System.err.println(deepToString(os));
	}
	static void fill(int[][] a, int val) {
		for(int i = 0; i < a.length; i++) Arrays.fill(a[i], val);
	}
	static void fill(int[][][] a, int val) {
		for(int i = 0; i < a.length; i++) fill(a[i], val);
	}

	public static void main(String[] args) throws Exception {
		new Main().run();
	}

	MyScanner sc = null;
	PrintWriter out = null;
	public void run() throws Exception {
		sc = new MyScanner(System.in);
		out = new PrintWriter(System.out);
		for (;sc.hasNext();) {
			solve();
			out.flush();
		}
		out.close();
	}

	class MyScanner {
		String line;
		BufferedReader reader;
		StringTokenizer tokenizer;

		public MyScanner(InputStream stream) {
			reader = new BufferedReader(new InputStreamReader(stream));
			tokenizer = null;
		}
		public void eat() {
			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
				try {
					line = reader.readLine();
					if (line == null) {
						tokenizer = null;
						return;
					}
					tokenizer = new StringTokenizer(line);
				} catch (IOException e) {
					throw new RuntimeException(e);
				}
			}
		}
		public String next() {
			eat();
			return tokenizer.nextToken();
		}
		public String nextLine() {
			try {
				return reader.readLine();
			} catch (IOException e) {
				throw new RuntimeException(e);
			}
		}
		public boolean hasNext() {
			eat();
			return (tokenizer != null && tokenizer.hasMoreElements());
		}
		public int nextInt() {
			return Integer.parseInt(next());
		}
		public long nextLong() {
			return Long.parseLong(next());
		}
		public double nextDouble() {
			return Double.parseDouble(next());
		}
		public int[] nextIntArray(int n) {
			int[] a = new int[n];
			for (int i = 0; i < n; i++) a[i] = nextInt();
			return a;
		}
	}
}