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.util.Scanner;

public class CFB {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);

		int n = sc.nextInt();
		int k = sc.nextInt();
		if (n == k) {
			System.out.println(-1);
			return ;
		}
		else if(k==n-1)
		{
			for(int i=1;i<=n;i++)
			{
				System.out.print(i);
				if(i==n)
					System.out.println();
				else
					System.out.print(" ");
			}
			return ;
		}
		System.out.print(n+" ");
		for (int i = 1,g =k+2, c = 0, j = 2; i <n; i++) {
			if (c< k)
				System.out.print(j++);
			else if(c==k)
				System.out.print(1);
			else
				System.out.print(g++);
			if(i==n-1)
				System.out.println();
			else
				System.out.print(" ");
			c++;
		}
	}

}