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.Scanner;

/**
 *
 * @author User
 */
public class A_Array_Peaks {
    
    public static void main(String[] args) {
        
        Scanner scan = new Scanner(System.in);
        
        int t = scan.nextInt();
        int n = 0;
        int k = 0;
        int[] ar;
        int[] ak;
        int test = 0;
        int swap = 0;
        int h = 1;
        int h1 = 1;
        int count = 0;
        
        
        for(int i=1;i<=t;i++){
            
            n = scan.nextInt();
            k = scan.nextInt();
            ak = new int[k+1];
            ar = new int[n+1];
            if((double)n/k<=2) System.out.println("-1");
            else{
                
                for(int j=2, y=n ; j<=n; j += 2){
                    
                    
                    if(k==0) break;
                    
                    ar[j] = y--;
                    k--;
                    
                }
                
                for(int j=1, w=1; j<=n; j++){
                    
                    if(ar[j]==0) ar[j] = w++;
                    
                }
                
                
                
                for(int j=1; j<n; j++){
                    
                    System.out.print(ar[j]+" ");
                    
                }
                System.out.println(ar[n]);
              
            }

   
        }
        
    }
    
}