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.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
public class LevkoAndPermutation {
	public static void main(String[]args)throws IOException{
		BufferedReader bf=new BufferedReader(new InputStreamReader(System.in));
		String s=bf.readLine();
		String[]sa=s.split(" ");
		int n=Integer.parseInt(sa[0]);
		int k=Integer.parseInt(sa[1]);
		if(k>n-1)
			System.out.println(-1);
		else{
			StringBuilder sb=new StringBuilder();
			for(int i=1;i<=n;i++){
				if(i<=n-k)
					if((n-k)%2==0){
						if(i%2==1)
							sb.append(" "+(i+1));
						else
							sb.append(" "+(i-1));
					}else{
						if(i==1)
							sb.append(" "+1);
						else{
							if(i%2==1)
								sb.append(" "+(i-1));
							else
								sb.append(" "+(i+1));
						}
					}
				else
					sb.append(" "+i);
			}
			System.out.println(sb.toString().substring(1));
		}
	}
}