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 CPP):

Solution:
#include <iostream>

using namespace std;

int main()
{
    int test;
    cin >> test;

    for(int t = 0; t < test; t++)
    {
        int n, k;
        cin >> n >> k;

        if(k < (n + 1)/2)
        {
            int arr[n];
            for(int i = 0; i < n; i++)
            {
                arr[i] = 0;
            }
            for(int j = 0; j < k; j++)
            {
                arr[2*j + 1] = n - j;
            }
            int y = 1;
            for(int l = 0; l < n; l++)
            {
                if(arr[l] == 0)
                {
                    arr[l] = y;
                    y++;
                }
            }
            for(int x = 0; x < n - 1; x++)
            {
                cout << arr[x] << " ";
            }
            cout << arr[n - 1] << endl;
        }
        else
        {
            cout << -1 << endl;
        }
    }

    return 0;
}