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>
#define pb push_back
#define ff first
#define ss second
#define ull unsigned long long
#define ll long long
#define int long long
#define pii pair <int, int>
#define all(v) v.begin(), v.end()
#define rall(v) v.rbegin(), v.rend()
#define rev(a) reverse(a.begin(), a.end())
#define endl '\n'
using namespace std;
const int inf = 1e18;
const int mod = 1e9 + 7;

void solve(){
	int n, k;
	cin >> n >> k;
	if (k > (n - 1) / 2){
		cout << -1;
		return;
	}
	set <int> s;
	for (int i = 1; i <= n; ++i) s.insert(i);
	vector <int> ans;
	int x = 1, y = n;
	for (int i = 0; i < k; ++i){
		ans.pb(x);
		s.erase(x++);
		if (s.size() > 0){
			ans.pb(y);
			s.erase(y--);
		}
	}
	for (auto i : s) ans.pb(i);
	for (auto i : ans) cout << i << " ";
}

signed main(){
	ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	int tt = 1;
	cin >> tt;
	while (tt--){
		solve();
		cout << endl;
	}
}