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>
using namespace std;

#define fo(i,n) for(int i=0;i<n;i++)
#define Fo(i,k,n) for(i=k;k<n?i<n:i>n;k<n?i+=1:i-=1)
#define ll long long
#define l long
#define PI 3.1415926535897932384626
#define nd "\n"

int main() {


std::ios::sync_with_stdio(false);

    int t = 1;
    cin >> t;
    int n,k;
    int possible_peaks;
    

    while (t--)
    {
       
        cin>>n>>k;
        int *arr = (int *)(malloc(n*4));
        if(n%2 == 0){
              possible_peaks = n/2-1;
        }
        else{
            possible_peaks = n/2;
        }

        if(k>possible_peaks){
            cout<<"-1"<<endl;
        }
        else{
           int  val = n;
           int i = 0;
           int pa = 1;
           while(val>n-k){
               if(i%2 ==0){
                   arr[i] = pa;
                   pa++;
                   i++;
               }
               else{
                   arr[i] = val;
                   val--;
                   i++;
               }
           }
           for(i;i<n;i++){
               arr[i] = pa;
               pa++;
           }
              fo(i,n){
                  cout<<arr[i]<<" ";
              }
              cout<<endl;
        }

        free(arr);
    }

    return 0;

}