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.IOException;
import java.io.InputStreamReader;

public class Main {
	
	public static void main(String[] args) throws NumberFormatException, IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		
		final String[] input = br.readLine().split("\\s+");
		final int n = Integer.parseInt(input[0]);
		final int k = Integer.parseInt(input[1]);
		
		/*
		 * 
		 */
		if (n == k) {
			System.out.println(-1);
		} else {
			StringBuilder sb = new StringBuilder();
			final String space = " ";
			int numBad = n - k;
			
			for (int i = 1; i <= numBad; ++i) {
				sb.append(i % numBad + 1).append(space);
			}
			
			for (int i = numBad + 1; i <= n; ++i) {
				sb.append(i).append(space);
			}
			
			System.out.println(sb.toString());
		}
	}
}