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>
#define ll long long
using namespace std;

int main()
{
	ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);

	int t;
	cin >> t;
	while(t--){
       int n, k;
       cin >> n >> k;
       if(n == 1 and k == 0){
       	 cout << 1 << "\n";
       	 continue;
       }
       if(n <= 2 and k > 0){
       	 cout << "-1\n";
       	 continue;
       }
       if(k > (n-1)/2){
       	 cout << "-1\n";
       	 continue;
       }
       vector<int> v;
       bool a[n+1];
       for(int i = 1; i <= n ;++i){
       	 a[i] = false;
       }
       for(int i = 1; i <= k; ++i){
       	  v.push_back(i);
       	  a[i] = true;
       	  v.push_back(n - i + 1);
       	  a[n - i + 1] = true;
       }
       for(int i = 1; i <= n; ++i){
       	 if(a[i] == false){
       	 	v.push_back(i);
       	 }
       }
       for(int i = 0; i < v.size(); ++i){
       	 cout << v[i] << " ";
       }
       cout << "\n";
	}
	return 0;
}