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;
int ar[110];
int main() {
int t;
cin>>t;
while(t-- ){ 
	int n ,k;
	cin>>n>>k; 
	memset(ar,0,sizeof ar);
	if(k==0){ 
		for(int i=1 ;i<=n ;i++ )cout<<i<<" ";
		cout<<endl;
	} 
	else { 
	if((n+1)/2<=k)cout<<-1<<endl;
else {  int j=2;
	for(int i=1 ;i<=n ; i++ ){ 
	if(k&&ar[i]==0){ cout<<i<<" "<<n-i+1<<" ";
	ar[n-i+1]=1;
	j++;
	k--;
	} 
	else if(ar[i]==0)cout<<i<<" ";
	}cout<<endl;
} 
} 
}
	
	return 0;
}