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 R210_Div2_B  //Name: Levko and Permutation
{
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int k = sc.nextInt();
		
		 StringBuilder sb = new StringBuilder();
         if (k == n)
         {
             sb.append(-1);
         }
         else        	 
         {
	         sb.append((n - k) + " ");
	         for (int i = 1; i < n - k; i++)
	         {
	             sb.append(i + " ");
	         }
	         if (k == 0)
	             sb.deleteCharAt(sb.length() - 1);
	         for (int i = n-k + 1; i <= n; i++)
	         {
	             sb.append(i);
	             if (i < n)
	                 sb.append(" ");
	         }
         }
		
		System.out.println(sb);
		sc.close();
	}
}
