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.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Scanner;

public class TaskB
{
    Scanner in;
    PrintWriter out;

    int n, k;
    ArrayList<Integer> answer = new ArrayList<Integer>();

    public static void main(String[] args)
    {
        TaskB mainTest = new TaskB();

        mainTest.readData();
        mainTest.solve();
        mainTest.outputData();
    }

    public TaskB()
    {
        in = new Scanner(System.in);
        out = new PrintWriter(new OutputStreamWriter(System.out));
    }

    public void readData()
    {
        n = in.nextInt();
        k = in.nextInt();
    }

    public void solve()
    {
        if (k < n) {
            for (int i = 0; i < n; i++) {
                if (i < n - k) {
                    if (i == 0) {
                        answer.add(n - k);
                    } else {
                        answer.add(i);
                    }
                } else {
                    answer.add(i + 1);
                }
            }
        } else {
            answer.add(-1);
        }
    }

    public void outputData()
    {
        for (int i = 0; i < answer.size(); i++) {
            out.print(answer.get(i));
            if (i + 1 == n) {
                out.println();
            } else {
                out.print(" ");
            }
        }
        out.flush();
    }
}