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>

#pragma GCC optimize("O3")

using namespace std;

int main(){
	cin.tie(0);
	ios::sync_with_stdio(0);
	int t;
	cin >> t;
	while(t--){
		int n, k;
		cin >> n >> k;
		if(k > (n-1)/2) cout << "-1\n";
		else{
			int l = 1, h = n, t = 1;
			cout << l++ << " ";
			while(k){
				if(t) cout << h-- << " ";
				else cout << l++ << " ";
				if(t) k--;
				t ^= 1;
			}
			if(t) for(int i = l; i <= h; i++) cout << i << " ";
			else for(int i = h; i >= l; i--) cout << i << " ";
			cout << "\n";
		}
	}
}