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 <iostream>
#include <vector>
#include <algorithm>
#include <queue>
#include <ctime>
#include <cmath>
#include <string>
#include <set>
#include <map>
#include <fstream>
#include <stack>

using namespace std;

typedef long long ll;
#define all(x) x.begin(), x.end()
//#define fi first
//#define se second
int inf = 1e9 + 7;



int main(){
    //srand(time(NULL));

    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    //*/
    /*
    cout.setf(ios::fixed);
    cout.precision(10);
    */
    /*
    ifstream fin;
    ofstream fout;
    fin.open("r.txt");
    fout.open("w.txt");
    */

   int t;
   cin >> t;
    for (int ctr = 0; ctr < t; ctr++){
        int n, k;
        cin >> n >> k;
        if (k > n / 2 + n % 2 - 1){
            cout << -1 << endl;
        }else{
            vector<int> ans(n + 10);
            ans[0] = 1;
            vector<bool> used(n + 10);
            int cnt = 3;
            for (int i = 0; i < k; i++){
                ans[i * 2 + 1] = cnt;
                used[cnt] = 1;
                cnt+= 2;
            }
            cnt = 2;
            for (int i = 0; i < k; i++){
                ans[i * 2 + 2] = cnt;
                cnt+= 2;
            }

            for (int i = k * 2 + 1; i < n; i++){
                ans[i] = cnt;
                cnt++;
            }
            for (int i = 0; i < n; i++){
                cout << ans[i] << ' ';
            }
            cout << endl;
        }
    }
    //cout << ans << endl;
    //cout << "NO" << endl;
    //fout.close();
    return 0;
}