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 <fstream>
#include <vector>
#include <set>
#include <algorithm>
#include <cmath>

using namespace std;

void solve(){
    int n,k;
    cin >> n >> k;
    if((int)(n-1)/2 < k){
        cout << -1;
    } else {
        int peaks = 0;
        int i = 1;
        for(; i <= n;i++){
            if(peaks == k){
                break;
            } else {
                int out = n-i+1;
                cout << i << " " << out << " ";
                peaks++;
            }
        }
        int I = i;
        for(; i <= n - I +1;i++){
            cout << i << " ";
        }
    }
    
}
  
int main() {

    // freopen("in.txt", "r", stdin);
    // freopen(".in", "r", stdin);
	// freopen(".out", "w", stdout);
   int N;
   cin >> N;
   for(int i = 0; i < N;i++){
       solve();
       cout << endl;
   }

    
}