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 peaks
{
    public static FastScanner stdin = new FastScanner();
    public static PrintWriter stdout = new PrintWriter(new BufferedOutputStream(System.out));


    public static void main(String[] args)
    {
        peaks peaks = new peaks();
    }
    public peaks()
    {
        // Fetching input
        int cases = stdin.nextInt();

        for (int i = 0; i < cases; i++)
        {
            int n = stdin.nextInt();
            int k = stdin.nextInt();

            if (n < 2*k+1)
                stdout.print("-1");
            else if(k==0)
            {
                for(int j = 1; j <= n; j++)
                {
                    stdout.print(j + " ");
                }
            }
            else if (k > 0)
            {
                int max = n;

                int arr[] = new int[n];
                for(int j = 0; j < k; j++,max--)
                {
                    arr[j*2+1] = max;
                }
                for(int j = 0, l = 1; j < n; j++)
                {
                    
                    if(arr[j]==0)
                    {
                        arr[j] = l;
                        l++;
                    }
                    stdout.print(arr[j] + " ");
                }
            }
            stdout.println();
            
        }


        // Sending output
        stdout.close();
    }
    private int solve() 
    {

        return 0;
    }

    /**
     * FastScanner Class
     * User-Defined to make I/O a bit faster
     */
    static class FastScanner 
    {
        BufferedReader br;
        StringTokenizer st;
        
        public FastScanner ()
        {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        String next()
        {
            while (st == null || !st.hasMoreElements())
            {
                try 
                {
                    st = new StringTokenizer(br.readLine());
                } 
                catch (IOException e)
                {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt()
        {
            return Integer.parseInt(next());
        }

        long nextLong()
        {
            return Long.parseLong(next());
        }

        double nextDouble()
        {
            return Double.parseDouble(next());
        }

        String nextLine()
        {
            String str = "";

            try 
            {
                str = br.readLine();
            }
            catch (IOException e)
            {
                e.printStackTrace();
            }

            return str;
        }
    }
}