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.Arrays;
import java.util.Scanner;

public class solution {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int f = scanner.nextInt();
        int[] array_n = new int[f];
        int[] array_k = new int[f];
        for (int i = 0; i < f; i++) {
            int n = scanner.nextInt();
            int k = scanner.nextInt();
            array_n[i] = n;
            array_k[i] = k;
        }
        for (int i = 0; i < f; i++) {
            int[] res = cs(array_n[i],array_k[i]);
            if (res == null)
                System.out.println(-1);
            else {
                for (int j = 0; j < res.length; j++) {
                    System.out.print(res[j] + " ");
                }
                System.out.println();
            }
        }
    }

    public static int[] cs(int n, int k) {
        int[] res =  new int[n];
        for (int i = 0; i < n; i++) {
            res[i] = i + 1;
        }
        if (k == 0) {
            return res;
        }
        else if (2*k + 1 > n) {
            return null;
        } else {
            for (int i = 1; i <= k ; i++) {
                int target = res[2*i];
                res[2*i] = res[2*i - 1];
                res[2*i - 1] = target;
            }
            return res;
        }
    }
}