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;
#define ll               long long int
#define pb                push_back
#define ppb               pop_back
#define pf                push_front
#define ppf               pop_front
#define all(x)            (x).begin(),(x).end()
#define uniq(v)           (v).erase(unique(all(v)),(v).end())
#define sz(x)             (int)((x).size())
#define fr                first
#define sc                second
#define pii               pair<int,int>
#define rep(i,a,b)        for(int i=a;i<b;i++)
#define mem1(a)           memset(a,-1,sizeof(a))
#define mem0(a)           memset(a,0,sizeof(a))
#define ppc               __sbuiltin_popcount
#define ppcll             __builtin_popcountll
#define IN                1000000000  
 
const int mod = 1000000007;    


int main()
{
    int t;
    cin >> t;
    while(t--)
    {
    	int n, k;
    	cin >> n >> k;
    	int a[n];
    	if(k > (n-1) / 2) 
		{
			cout << -1 << "\n";
		}
    	else
    	{
    	for(int i=0;i<n;i++) a[i] = -1;
    	
    	int last = n;
    	for(int i=1;i<n;i+=2)
    	{
    		if(k == 0) break;
    		a[i] = last;
    		last--;
    		k--;
		}
		
		for(int i=0;i<n;i++)
		{
			if(a[i] == -1)
			{
				a[i] = last;
				last--;
			}
		}
		for(int i=0;i<n;i++)
		cout << a[i] << " ";
		cout << "\n";
    	} 
	}
	return 0;
}