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;

void permutation(int n, int k)
{
    int i = n-k+1,j=(n-k)/(k+1),a=1,b;

    while(i<=n)
    {
        b=a+j;
        for(a;a<b;a++)
        cout<<a<<" ";
        cout<<i<<" ";
        i++;
    }
    for(a;a<=n-k;a++)
    cout<<a<<" ";
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int t;
    cin>>t;
    while(t--)
    {
        int n,k;
        cin>>n>>k;
        
        if(k>(n-1)/2)
        cout<<"-1";
        else if(k==0)
        {
            for(int i=1;i<=n;i++)
            cout<<i<<" ";
        }
        else
        permutation(n,k);
        cout<<endl;
    }
    return 0;
}