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 CodeForces1513A{
	public static void main(String[] args) {
		Scanner input = new Scanner(System.in);
		int t = input.nextInt();
		for(int j = 0;j<t;j++){
			int n = input.nextInt();
			int k = input.nextInt();
			if(k > n/2 || (k == n/2 && n%2 == 0)){
				System.out.println(-1);
			}
			else{
				int x = 1;
				boolean bool = true;
				while(k > 0){
					if(bool){
						System.out.print(x + " ");
						bool = false;	
						x++;
					}
					else{
						System.out.print(n + " ");
						n--;
						bool = true;
						k--;
					}
				}

				for(int i = x;i<=n;i++){
					System.out.print(i + " ");
				}

				System.out.println();

			}
		}
	}
}