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;

/**
 * @author Wangmingcan
 * @date 2021/5/3 10:32
 * @description
 */
public class B_ArrayAndPeaks {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while (t-- > 0) {
            int n = sc.nextInt();
            int k = sc.nextInt();
            if (n < 3 && k!= 0) {
                System.out.println(-1);
                continue;
            }
            if (k >= (n+1)/2) {
                System.out.println(-1);
                continue;
            }
            int[] arr = new int[n];
            for (int i = 0; i < n;i++) {
                arr[i] = i+1;
            }
            if (k==0) {
                for (int i = 0; i < n;i++) {
                    System.out.print(arr[i] + " ");
                }
                System.out.println();
                continue;
            }
            for (int i = 0; k > 0;) {
                int temp = arr[n-1-2*i-1];
                arr[n-1-2*i-1] = arr[n-1-2*i];
                arr[n-1-2*i] = temp;
                k--;
                i++;
            }
            for (int i = 0; i < n;i++) {
                System.out.print(arr[i] + " ");
            }
            System.out.println();
        }
    }
}