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:
// #pragma GCC optimize(2)
// #pragma GCC optimize(3)
// #pragma GCC optimize("Ofast")
#include<bits/stdc++.h>
using namespace std;
#define _rep(i, x, y) for(int i = (int)x; i < (int)y; ++i)
#define _dep(i,x,y) for(int i = (int)x; i > (int)y; i--)
#define PII pair<int,int>
#define eb emplace_back
#define pb push_back
#define fi first
#define se second
#define PQ priority_queue
#define lb lower_bound
#define ub upper_bound
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
typedef long long ll;
typedef unsigned long long ull;
typedef vector<int> VI;
typedef vector<VI> VII;
typedef vector<ll> VL;
typedef vector<vector<ll>> VLL;
constexpr int mod = 1e9 + 7;
constexpr int KINF = 0x3f3f3f3f;
constexpr int INF = 0x7f7f7f7f;
constexpr double eps = 1e-7;
int main(){
	ios::sync_with_stdio(false); cin.tie(0), cout.tie(0);
	int _;
	cin >> _;
	while(_--){
		int n, k;
		cin >> n >> k;
		if(k > (n - 1) / 2) cout << -1 << endl;
		else{
			VI ans(n);
			iota(all(ans), 1);
			int i = 1;
			while(k){
				swap(ans[i], ans[i + 1]);
				i += 2;
				k--;
			}
			_rep(i, 0, n) cout << ans[i] << " \n"[i == n - 1];
		}
	}
	return 0;
} 
// 小心左移运算符导致的int溢出
// dp记得处理非法状态
// resize并不会初始化数组
// 全局函数一定要记得inline
// 带空格字符串不能直接cin,要用getline
// 最好用用"\n"代替endl