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 Main {
  public static void main (String[] args) {
	  Scanner in = new Scanner(System.in);
	  int n = in.nextInt();
	  int k = in.nextInt();
	  StringBuilder s = new StringBuilder();
	  if (n == k) {
		  s.append(-1);
	  } else if (n == k+1) {
		  for (int i = 1; i <= n; i++) {
			  s.append(i);
			  if (i < n) {
				  s.append(" ");
			  }
		  }
	  } else {
		  s.append(k+2);
		  for (int i = 1; i <= k; i++) {
			  s.append(" ");
			  s.append(i+1);
		  }
		  for (int i = k+1; i < n-1; i++) {
			  s.append(" ");
			  s.append(i+2);
		  }
		  s.append(" ");
		  s.append(1);
	  }
	  System.out.println(s);
	  in.close();
  }
}