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> 
#include <stdio.h>
#include <cstdlib>
#include<algorithm> 
#include<string.h>
#include<map>
#include<math.h>
using namespace std; 
int main()
{
int t;
cin>>t;
while(t--)
{
int n,k;
cin>>n>>k;int count =0;
vector <int> v,a(n);int max=n;
if(k>(n-1)/2)
 {   cout<<"-1"<<endl;
    continue;
}
for(int i=1;i<n;i+=2)
{
    if(count==k)
        break;
    a[i]=max;
    max--;
        count++;

}
for(int i=0;i<n;i++)
{
    if(a[i]!=0)
        continue;
    a[i]=max;
    max--;
}
for(int i=0;i<n;i++)
{
    cout<<a[i]<<" ";
}

cout<<endl;
}
}