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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class Round714Div2A {
	public static void main(String[] args) throws NumberFormatException, IOException {
		BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
		int numT = Integer.parseInt(f.readLine());
		for (int i = 0; i < numT; i++) {
			StringTokenizer st = new StringTokenizer(f.readLine());
			int n = Integer.parseInt(st.nextToken());
			int k = Integer.parseInt(st.nextToken());
			if(n-k<k+1) {
				System.out.println(-1);
				continue;
			}
			int s = 0;
			int l = 0;
			int[] results = new int[n];
			for (int j = 0; j < n; j++) {
				if(l<k) {
					if(j%2==1) {
						results[j]=n-l;
						l++;
					}else {
						results[j]=s+1;
						s++;
					}
				}else {
					results[j]=s+1;
					s++;
				}
			}
			for (int j = 0; j < results.length-1; j++) {
				System.out.print(results[j]+" ");
			}
			System.out.println(results[n-1]);
		}
	}
}