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>
#include <iomanip>
#include <algorithm>
#include <vector>
#include <cmath>
#include <queue>
#include <map>
#include <set>
#include <utility>
#include <math.h>
#include <string>
#include <cstring>
#include <cassert>
using namespace std;
 
const double PI = acos(-1);
const int N=1e5 + 10;
const int M=1000000007;
typedef long long ll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int, int> ii;
#define pb push_back
#define mp make_pair
#define f first
#define s second
#define all(c) (c).begin(),(c).end()
#define rall(c) (c).rbegin(),(c).rend()
#define oo 1000000001


void solve(){
	int n,k;
	cin>>n>>k;
	if(k>(n-1)/2){
		printf("-1\n");
		return;
	}
	int mn=1,mx=n;
	for (int i = 0; i < n; ++i)
	{
		if(k && i&1){
			k--;
			printf("%d ",mx-- );
		}else
			printf("%d ",mn++ );
	}
	printf("\n");
}
 
int main(){
	int t=1;
	scanf("%d",&t);
	while(t--)
		solve();

	return 0;
}