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 <bits/stdc++.h>
using namespace std;
typedef long long ll;
int n, k, a[120]; 
int main()
{
	ios::sync_with_stdio(false);
	int T;
	cin >> T;
	while(T--)
	{
		cin >> n >> k;
		if(k * 2 >= n) cout << -1 << endl;
		else
		{
			for(int p = 0, i = 1; p <= k; p++, i++) a[2 * p + 1] = i;
			for(int p = 1, j = k + 2; p <= k; p++, j++) a[2 * p] = j;
			for(int i = 2 * k + 2; i <= n; i++) a[i] = i;
			for(int i = 1; i <= n; i++)
			{
				cout << a[i];
				if(i == n) cout << endl;
				else cout << ' ';
			}
		} 
	}
	return 0;
}