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.util.Scanner;
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
 * @author balbasaur
 */
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();
        int n1 = k;
        if (n == k){
            out.print(-1);return;
        }
        boolean []al = new boolean[100002];
        int []res = new int[n + 1];
        int ii = 1;
        int count = 1;
        int i = 1;
        for (i = 2;k > 0 && i <= n; ++i){
            res[i] = i;k--;
            al[i] = true;
        }
        for (;i <= n;++i){
            res[i] = i == n?1:i + 1;
            if (i == n){
                al[1] = true;
            }
            else al[i + 1] = true;
        }
        int z = 1;
        while (al[z] == true)z++;
        res[1] = z;

        for (i = 1;i <=n; ++i){
            out.print(res[i]+" ");
        }

    }
}

class InputReader {
    public BufferedReader reader;
    public StringTokenizer tokenizer;

    public InputReader(InputStream stream) {
        reader = new BufferedReader(new InputStreamReader(stream));
        tokenizer = null;
    }

    public String next() {
        while (tokenizer == null || !tokenizer.hasMoreTokens()) {
            try {
                tokenizer = new StringTokenizer(reader.readLine());
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
        }
        return tokenizer.nextToken();
    }

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


}

