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.io.BufferedReader;
import java.io.InputStreamReader;


public class TaskB {
	
	public static void main(String[] args) throws Exception {
		BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
		String[] split = reader.readLine().split(" ");
		int n = Integer.valueOf(split[0]);
		int k = Integer.valueOf(split[1]);
		
		if (k <= n - 1) {
			int[] perm = new int[n+1];
			for (int i = 1; i <= n; i++) {
				perm[i] = i;
			}
			
			int idx = n;
			for (int i = 0; i < (n-1) - k; i++) {
				int temp = perm[idx];
				perm[idx] = perm[1];
				perm[1] = temp;
				idx--;
			}
			
			System.out.print(perm[1]);
			for (int i = 2; i <= n; i++) {
				System.out.print(" " + perm[i]);
			}
		} else {
			System.out.println("-1");
		}
	}
}
