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.*;

public class LevkoAndPermutation {
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		
		int n = in.nextInt();
		int k = in.nextInt();
		// Property: gcd(1, x) = 1, gcd(x, x + 1) = 1 -> shift first n-k elements 1 position
		// Property: k == n, impossible
		
		if (k == n) {
			System.out.println(-1);
		} else {
			int[] permutation = new int[n + 1];
			
			for (int i = 2; i <= n - k; ++i)
				permutation[i] = i - 1;
			permutation[1] = n - k;
			for (int i = n - k + 1; i <= n; ++i)
				permutation[i] = i;
			
			StringBuilder sb = new StringBuilder();
			for (int i = 1; i <= n; ++i) {
				if (i > 1) sb.append(" ");
				sb.append(permutation[i]);
			}
			System.out.println(sb);
		}
		
		in.close();
		System.exit(0);
	}
}