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 scanner = new Scanner(System.in);
		int n = scanner.nextInt();
		int k = scanner.nextInt();
		solver(n, k);
	}

	private static void solver(int n, int k) {
		if (n == k) {
			System.out.println(-1);
		} else {
			int[] data = new int[n];
			for (int i = 0; i < k; i++) {
				data[i + 1] = i + 2;
			}
			if (k + 1 < n) {
				data[k + 1] = 1;
			}

			data[0] = k == n - 1 ? 1 : n;

			for (int i = k + 2; i < n; i++) {
				data[i] = i;
			}
			for (int i = 0; i < n; i++) {
				System.out.print(String.format("%d ", data[i]));
			}
		}
	}
}