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.OutputStream;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import java.util.StringTokenizer;

public class Gym {

	static Map<Integer, Long> map = new HashMap<>();

	public static void main(String[] args) {
		FastScanner fs = new FastScanner();
		OutputStream outputStream = System.out;
		PrintWriter out = new PrintWriter(outputStream);
		int t = 1;// fs.nextInt();
		while (t-- > 0) {

			solve(fs, out);
		}
		out.flush();
		out.close();
	}

	private static void solve(FastScanner fs, PrintWriter out) {
		int n = fs.nextInt(), k = fs.nextInt();
		if (k == n) {
			System.out.println(-1);
		} else {
			int a[] = new int[n];
			if (k == n - 1) {
				for (int i = 0; i < a.length; i++) {
					a[i] = i + 1;
				}
			} else {
				a[0] = n - k;
				for (int i = 1; i < n - k; i++) {
					a[i] = i;
				}
				for (int i = n - k; i < a.length; i++) {
					a[i] = i + 1;
				}
			}
			for (int i = 0; i < a.length; i++) {
				out.print(a[i] + " ");
			}
		}
	}

	static boolean isPrime(long n) {
		if (n <= 1)
			return false;
		if (n <= 3)
			return true;
		if (n % 2 == 0 || n % 3 == 0)
			return false;
		for (int i = 5; i * i <= n; i = i + 6)
			if (n % i == 0 || n % (i + 2) == 0)
				return false;
		return true;
	}

	static void sieveOfEratosthenes(int n, int factors[]) {
		factors[1] = 1;
		for (int p = 2; p * p <= n; p++) {
			if (factors[p] == 0) {
				factors[p] = p;
				for (int i = p * p; i <= n; i += p)
					factors[i] = p;
			}
		}
	}

	static long gcd(long a, long b) {
		if (b == 0)
			return a;
		return gcd(b, a % b);
	}

	static class FastScanner {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer("");

		public String next() {
			while (!st.hasMoreElements())
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}

		int nextInt() {
			return Integer.parseInt(next());
		}

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

		long nextLong() {
			return Long.parseLong(next());
		}
	}
}
