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>
using namespace std;
#include<string.h>

int main() {
    int t;
    cin>>t;
    while(t--)
    {
        int count=0;
       int n,k;
       cin>>n>>k;
       if(k>=double(n)/2)
       {
           cout<<"-1"<<endl;
       }
       else
       {
       
       int arr[n];
       int c=n;
       for(int i=0;i<n;i++)
       {
           arr[i]=c;
           c--;

       }
       for(int i=0;i<n;i+=2)
       {
           if(count==k)
         {
             break;
         }
           int temp=arr[i];
         arr[i]=arr[i+1];
         arr[i+1]=temp;
         count++;
         
       }
       for(int i=0;i<n;i++)
       {
           cout<<arr[i]<<" ";
       }
       cout<<endl;
       }
       
    }
   
}