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 program3{
    public static void main(String[] args) {
        Scanner in=new Scanner(System.in);
       int t=in.nextInt();
       for (int ts=0;ts<t;ts++) {
    	   int n=in.nextInt();
    	   int k=in.nextInt();
    	   int [] arr=new int [n];
    	   if (k*2>=n) {
    		   System.out.println(-1);
    		   continue;
    	   }
    	  for (int i=0;i<n;i++) {
    		  arr[i]=i+1;
    	  }
    	  int index=1;
    	  for (int j=0;j<k;j++) {
    		 int temp=arr[index];
    		 arr[index]=arr[index+1];
    		 arr[index+1]=temp;
    		  index+=2;
    	  }
    	  
    	   for (int i=0;i<n;i++) {
    		   if (arr[i]==0) {
    			   arr[i]=i+1;
    			   
    		   }
    		   System.out.print(arr[i]+" ");
    	   }
    	   System.out.println();
       }
		
        in.close();
        
    }
}