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.*;
import java.util.StringTokenizer;

public class A {

	public void execute(int input) throws Exception {

		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

		if (input == 1) {
			br = new BufferedReader(new FileReader("test.txt"));
		}

		int nTest = Integer.parseInt(br.readLine());

		for (int t = 0; t < nTest; t++) {

			StringTokenizer st = new StringTokenizer(br.readLine());

			int n = Integer.parseInt(st.nextToken());
			int k = Integer.parseInt(st.nextToken());

			int[] a = new int[n];
			for (int i = 0; i < n; i++) {
				a[i] = i + 1;
			}

			int j = 1;
			while (k > 0 && j + 1 < n) {

				int temp = a[j];
				a[j] = a[j + 1];
				a[j + 1] = temp;

				j += 2;
				k--;
			}

			if (k == 0) {
				for (int i = 0; i < n - 1; i++) {
					System.out.print(a[i] + " ");
				}
				System.out.println(a[n - 1]);
			} else {
				System.out.println(-1);
			}
		}
	}

	public static void main(String[] args) throws Exception {

		A a = new A();

		a.execute(0);

	}
}