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.*;
public class ArrayandPeaks {
	private static StreamTokenizer st;
	private static int nextInt() throws IOException{
		st.nextToken();
		return (int)st.nval;
	}
	public static void main(String[] args) throws IOException{
		st = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
		int t = nextInt();
		PrintWriter pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
		for(int x = 0; x < t; ++x) {
			int n = nextInt(), k = nextInt();
			if(2*k>n-1)pw.println(-1);
			else {
				for(int i = 0; i < k; ++i) {
					pw.print((i+1)+" "+(n-i) + " ");
				}
				for(int i = k+1; i <= n-k; ++i) {
					pw.print(i + " ");
				}
				pw.println();
			}
		}
		pw.close();
	}

}