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;

public class LevkoandPermutation
{
	public static void main(String[] args)
	{
		Scanner input = new Scanner(System.in);
		int n = input.nextInt();
		int k = input.nextInt();
		if (n == k)
		{
			System.out.println(-1);
			return;
		}
		StringBuilder sb = new StringBuilder();
		int idx = -1;
		if ((n - k) % 2 == 0)
		{
			idx = 0;
		}
		else
		{
			sb.append("1 ");
			idx = 1;
		}
		for (; idx < n - k; idx += 2)
		{
			sb.append((idx + 2) + " " + (idx + 1) + " ");
		}
		for (idx = n - k + 1; idx <= n; ++idx)
		{
			sb.append(idx + " ");
		}
		sb.setCharAt(sb.length() - 1, '\n');
		System.out.print(sb);
	}
}