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;
import java.util.Arrays;
import java.util.HashMap;
//import java.util.HashMap;
//import java.util.Arrays;
//import java.util.Stack;
import java.lang.Math;
public class solution {
	
	
	                
	public static void main(String[] args) {
		Scanner s = new Scanner(System.in);
		int testcases = s.nextInt();
		while(testcases-->0) {
	
			
			int n=s.nextInt();
//			int[] arr=new int[n+2];
//			for(int i=0;i<n+2;i++) {
//				arr[i]=s.nextInt();
//			}
			
			int k=s.nextInt();
			
			if(k>(n-1)/2) {
				System.out.println(-1);
				continue;
			}
			
			int arr[]=new int[n];
			arr[0]=1;
			
			for(int i=1;i<n;) {
				
				if(k==0) {
					for(;i<n;i++) {
						arr[i]=i+1;
					}
					
					break;
				}
				else {
					arr[i]=i+2;
					arr[i+1]=i+1;
					k--;
					i+=2;
				}
				
			}
			
			for(int i=0;i<n;i++) {
				System.out.print(arr[i]+" ");
			}
			
			System.out.println();
			
			
	
		
		
		}
	

}
}