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<bits/stdc++.h>
using namespace std;

int main() {
  int t;
  cin>>t;
  while(t--)
  {
      int n,k;
      cin>>n>>k;
      int a[n];
      int x=1;
      for(int i=0;i<n;i++)
      {
          a[i]=x;
          x++;
      }
      if(n<2*k+1)
      cout<<-1<<endl;
      else
      {
          int i=1;
      while(k--)
      {
        
              //swap
              int temp=a[i];
              a[i]=a[i+1];
              a[i+1]=temp;
              i=i+2;
         
      }
       for(int i=0;i<n;i++)
      {
          cout<<a[i]<<" ";
      }
      
       cout<<endl;
      }
     
      
  }
}