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:
//stan hu tao
//come to K-expo!!!
//watch me get carried in nct ridin
import static java.lang.Math.max;
import static java.lang.Math.min;
import static java.lang.Math.abs;
import static java.lang.System.out;
import java.util.*;
import java.io.*;
import java.math.*;

public class x1513A
{
    public static void main(String hi[]) throws Exception
    {
        BufferedReader infile = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(infile.readLine());
        int T = Integer.parseInt(st.nextToken());
        StringBuilder sb = new StringBuilder();
        while(T-->0)
        {
            st = new StringTokenizer(infile.readLine());
            int N = Integer.parseInt(st.nextToken());
            int K = Integer.parseInt(st.nextToken());
            ArrayDeque<Integer> q = new ArrayDeque<Integer>();
            for(int v=2; v <= N; v++)
                q.add(v);
            int[] res = new int[N];
            res[0] = 1;
            for(int i=1; i < N; i++)
            {
                if(K > 0)
                {
                    if(i%2 == 1)
                        res[i] = q.pollLast();
                    else
                    {
                        res[i] = q.pollFirst();
                        K--;
                    }
                }
                else
                    res[i] = q.pollFirst();
            }
            if(K > 0)
                sb.append("-1\n");
            else
            {
                for(int x: res)
                    sb.append(x+" ");
                sb.append("\n");
            }
        }
        System.out.print(sb);
    }
    public static int[] readArr(int N, BufferedReader infile, StringTokenizer st) throws Exception
    {
        int[] arr = new int[N];
        st = new StringTokenizer(infile.readLine());
        for(int i=0; i < N; i++)
            arr[i] = Integer.parseInt(st.nextToken());
        return arr;
    }
}