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.Scanner;

public class ArrayAndPeek {
    private static void solve(int n, int k) {
        if (k > (n - 1) / 2) {
            System.out.println(-1);
            return;
        }

        int nonPeek = 1;

        for (int peek = n - k + 1; nonPeek <= n - k && peek <= n; nonPeek++, peek++) {
            System.out.print(nonPeek + " " + peek + " ");
        }

        for (; nonPeek <= n - k; nonPeek++) {
            System.out.print(nonPeek + " ");
        }

        System.out.println();
    }

    public static void main(String[] args) {

        Scanner scanner = new Scanner(System.in);

        int cases = scanner.nextInt();
        int[][] inputs = new int[cases][2];
        for (int i = 0; i < cases; i++) {
            inputs[i][0] = scanner.nextInt();
            inputs[i][1] = scanner.nextInt();
        }

        for (int[] input : inputs) {
            solve(input[0], input[1]);
        }
    }
}