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;

const int N = 100010;
int ans[N];

int main()
{
    int q;
    cin >> q;
    while(q--){
        memset(ans, 0 ,sizeof ans);
        int n, k;
        cin >> n>> k;
        if(k > (n - 1) / 2)cout <<-1<<endl;
        else {
                int cnt = k;
                int tt =0;
            for(int i = n; cnt > 0 ; i--){
                ans[2 * (++tt)] = i;
                cnt --;
            }
            tt = 0;
            for(int i = 1; i <= n; i++)
            {
                if(ans[i])continue;
                else {
                    ans[i] = ++tt;
                }
            }
            for(int i = 1;i <=n; i++)cout <<ans[i]<<" ";
            cout <<endl;
        }

    }
}