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 all(x) (x).begin(), (x).end()
#define endl "\n"
#define pb(x) push_back(x)
#define fio() ios_base::sync_with_stdio(0); cin.tie(0)
#define fileio() ifstream file_in("boards.txt");ofstream file_out("boards.txt")
#define f first
#define s second
//#define double long double
#define db double
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef tuple<int, int, int> tpi;
typedef tuple<ll, ll, ll> tpl;
typedef pair<double, ll> pdl;
typedef pair<double, int> pdi;
typedef pair<double, double> pdd;
const double pi = acos(-1);
const double eps = 1e-10;
const int dx[] = { 0,1,0,-1 };
const int dy[] = { 1,0,-1,0 };
const int MOD = 1e9+7;
const ll LMOD = 1e9+7; // PLZ check
const int INF = 0x3f3f3f3f;
const ll LINF = 0x3f3f3f3f3f3f3f3f;

void solve() {
	int n, k;
	cin >> n >> k;
	if (k > (n-1)/2) {
		cout << -1 << endl;
		return;
	}

	vector<int> a(n);
	for (int i = 0; i < n; i++) a[i] = i+1;
	
	int p = 1;
	while (p < n && k--) {
		swap(a[p], a[p+1]);
		p += 2;
	}

	for (int x : a) cout << x << ' ';
	cout << endl;
}
int main() {
	fio();
	int t;
	cin >> t;
	while (t--) solve();
	return 0;
}