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 B {
	public static void main(String[] args){
		Scanner input = new Scanner(System.in);
		int n = input.nextInt();
		int k = input.nextInt();
		if (k >= n) System.out.println(-1);
		else if ((n-k) % 2 == 1){
			System.out.print(1);
			for (int i=2; i<=k+1; i++)
				System.out.print(" "+i);
			for (int i=k+2; i<=n; i+=2)
				System.out.print(" "+(i+1)+" "+i);
			System.out.println();
		}
		else{
			System.out.print(2+" "+1);
			for (int i=3; i<=k+2; i++)
				System.out.print(" "+i);
			for (int i=k+3; i<=n; i+=2)
				System.out.print(" "+(i+1)+" "+i);
			System.out.println();
		}
	}
}