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 fast_cin() ios_base::sync_with_stdio(false); cin.tie(NULL) ;cout.tie(NULL)
using namespace std;

#define ll long long int
#define pb push_back
#define endn "\n"
#define mp make_pair
#define ss second
#define ff first
#define deb(x) cout << #x << " " << x << endl
#define deb1(x,y) cout<<#x<<" "<<x<<" , "<<#y<<" "<<y<<endl
#define deb2(x,y,z) cout<<#x<<" "<<x<<" , "<<#y<<" "<<y<<" , "<<#z<<" "<<z<<endl
const long long INF = 1e18;

int small = INT_MIN;
int large = INT_MAX;

void solve() {
	int n, k;
	cin >> n >> k;
	int a[n];
	int i = 2, j = n;
	a[0] = 1;
	int t = 1;

	while (k-- && t < n - 1) {
		a[t] = j;
		t++;
		j--;
		a[t] = i;
		t++;
		i++;
	}

	if (k != -1) {
		cout << -1;
	}
	else {
		while (i <= j) {
			a[t] = i;
			t++;
			i++;
		}
		for (int x : a) {
			cout << x << " ";
		}
	}
	cout << endn;

}

int main() {

	fast_cin();

	ll T = 1;
	cin >> T;
	while (T--) solve();

	return 0;
}