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 <sstream>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <cstdio>
#include <cstdlib>
#include <cctype>
#include <complex>
#include <malloc.h>
#include <cmath>
#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
#include <bitset>
#include <list>
#include <string.h>
#include <assert.h>
#include  <time.h>
using namespace std;

#define ll long long
#define lp(i, e, n) for(ll i = e; i < n; i++)
#define IO ios::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)
//inevitability is the unavailabilty of vulnerability.
typedef int int32_t;
typedef pair<int, int> ii;
vector<int> v;

const char nl = '\n';
const int maxn = 2e5 + 17;

void ontheGround(){
	int t; cin >> t;
	lp(i, 0, t){
		int n, k; cin >> n >> k;
	int beak[n]; int cnt{0};
	int max = (n - 1) / 2;
	if(k > max){
		cout << "-1" << "\n";
		continue;
	}
	map <int, int> mp;
        int s=1,e=n;
 
        int lim=k*2+1;
        for(int i = 1; i<=n && lim; i++, lim--)
        {
            if(i%2 == 0)
            {
                cout<<e<<" ";
                e--;
            }
            else
            {
                cout << s << " ";
                s++; 
            }
        }
        for (int i = s; i<=e; i++)
        {
            cout<< i << " ";
        }
	cout << endl;
	}
} 

/* 
no thats not me.
*/
int main(){
	//freopen("input.txt", "r", stdin);
	//freopen("output.txt", "w", stdout);
	IO;
	ll TEST = 1;
	//cin >> TEST;
	lp(i, 0, TEST){
		ontheGround();
	}

return 0; 
}