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.*;
  import java.io.*;
  
  public class Main {
    public static void main(String args[]) throws IOException {
       Scanner sc = new Scanner(System.in);
      int t =sc.nextInt();
      while(t-->0)
      {  int n =sc.nextInt(),n1=n;
          int k =sc.nextInt(),k1=k;
          if(n%2==0&&k>(n/2)-1){
          System.out.println("-1");
          continue;}
          if(n%2!=0&&k>n/2){
          System.out.println("-1");
          continue;}
          for(int i=1;i<=n-k;i++)
          { System.out.print(i+" ");
            if(k1>0)
            { System.out.print(n1--+" ");
               k1--;
            }
          }
          System.out.println();
      }
    }
  }