Problem: A sequence of n integers is called a permutation if it contains all integers from 1 to n exactly once.

Given two integers n and k, construct a permutation a of numbers from 1 to n which has exactly k peaks. An index i of an array a of size n is said to be a peak if 1 < i < n and a_i \gt a_{i-1} and a_i \gt a_{i+1}. If such permutation is not possible, then print -1.

Input

The first line contains an integer t (1 ≤ t ≤ 100) — the number of test cases.

Then t lines follow, each containing two space-separated integers n (1 ≤ n ≤ 100) and k (0 ≤ k ≤ n) — the length of an array and the required number of peaks.

Output

Output t lines. For each test case, if there is no permutation with given length and number of peaks, then print -1. Otherwise print a line containing n space-separated integers which forms a permutation of numbers from 1 to n and contains exactly k peaks. 

If there are multiple answers, print any.

Example

Input


5
1 0
5 2
6 6
2 1
6 1


Output


1 
2 4 1 5 3 
-1
-1
1 3 6 5 4 2

Note

In the second test case of the example, we have array a = [2,4,1,5,3]. Here, indices i=2 and i=4 are the peaks of the array. This is because (a_{2} \gt a_{1} , a_{2} \gt a_{3}) and (a_{4} \gt a_{3}, a_{4} \gt a_{5}). 

Solution (in JAVA):

Solution:
import java.util.*;
public class a {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner sc = new Scanner(System.in);
		int t = sc.nextInt();
		
		while(t-->0)
		{
			int n = sc.nextInt();
			int k = sc.nextInt();
			
			TreeSet<Integer> tr = new TreeSet<>();
			if((n%2==1 && k>n/2) ||(n%2==0 && k>=n/2))
			{
				System.out.println(-1);
				continue;
			}
			for(int i = 1; i <=n; i++)
			{
				tr.add(i);
			}
			
			
			for(int i = 0; i < k; i++)
			{
				System.out.print(tr.pollFirst()+" "+tr.pollLast()+" ");
				
			}
			int len = tr.size();
			for(int i=0; i < len; i++)
			{
				System.out.print(tr.pollFirst()+" ");
			}
			
			System.out.println();
		}
	}

}