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.io.*;
import java.util.*;

public class problem {

	public static void main(String[] args)throws IOException {
		// TODO Auto-generated method stub
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		BufferedWriter out = new BufferedWriter(new OutputStreamWriter(System.out));
		int t = Integer.parseInt(br.readLine());
		for(int i=0; i<t; i++) {
			StringTokenizer st = new StringTokenizer(br.readLine());
			int N = Integer.parseInt(st.nextToken());
			int M = Integer.parseInt(st.nextToken());
			int max = (N-1)/2;
			if(M>max) {
				out.write(-1+" ");
				out.write("\n");
			}
			else {
				int[] arr = new int[N];
				for(int j=0; j<N; j++)arr[j] = j+1;
				int j = 1;
				while(M!=0) {
					int save = arr[j];
					arr[j] = arr[j+1];
					arr[j+1] = save;
					j+=2;
					M--;
				}
				for(j=0; j<N; j++)
					out.write(arr[j]+" ");
				out.write("\n");
			}
		}
		out.flush();
		out.close();
	}

}