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 LevkoPermutation
{
	public static void main(String[] args)
	{
		Scanner in = new Scanner(System.in);
		int N = in.nextInt();
		int K = in.nextInt();
		in.close();

		StringBuilder sb = new StringBuilder();
		if (K == N)
		{
			sb.append(-1);
		}
		else
		{
			int[] arr = new int[N + 1];
			for (int x = 1; x <= N; x++)
			{
				arr[x] = x;
			}

			int count = N - 1;
			int x = 2;
			while (count - 1 > K)
			{
				int temp = arr[x];
				arr[x] = arr[x + 1];
				arr[x + 1] = temp;

				x += 2;
				count -= 2;
			}

			if (count != K)
			{
				arr[x] = 1;
				arr[1] = x;
			}

			for (int c = 1; c <= N; c++)
			{
				sb.append(arr[c]).append(' ');
			}
			sb.deleteCharAt(sb.length() - 1);
		}
		System.out.println(sb);
	}

}
