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:
//#pragma GCC optimize("Ofast")
//#pragma GCC target("avx,avx2,fma")
//#pragma GCC optimization ("unroll-loops")

#include <bits/stdc++.h>
using namespace std;
using namespace chrono;

#define ll long long
#define ld long double
#define mod 1000000007
#define pb push_back
#define inf 1000000000000000000
#define ff first
#define ss second
#define deb(x) cout<<#x<<" "<<x<<"\n"
#define Clear(x) memset(x,0,sizeof(x))
#define all(x) (x).begin(),(x).end()

void checkpoint1()
{
	/******think more code less******/
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	#ifndef ONLINE_JUDGE
		freopen("input.txt","r",stdin);
		freopen("output.txt","w",stdout);
		freopen("Error.txt", "w", stderr);
	#endif
}

bool comp(pair< ll,ll > &a,pair< ll,ll > &b)
{
	if(a.ff!=b.ff)
		return a.ff<b.ff;
	else
	{
		if(a.ss>b.ss)
			return b.ss<a.ss;
		else if(a.ss<b.ss)
			return a.ss<b.ss;
	}
	return a.ff>b.ff;
}

ll powe(ll a,ll b)
{
	ll res=1;
	while(b>0)
	{
		if(b&1)
		{
			res*=a;
			res%=mod;
		}
		a*=a;
		a%=mod;
		b>>=1;
	}
	return res;
}

void terminAtor()
{
	ll t;
	cin>>t;
	while(t--)
	{
		ll n,k;
		cin>>n>>k;
		ll a[n+1];
		for(ll i=1;i<=n;i++)
			a[i]=i;
		ll nop=0,i=n;
		while(nop<k && i>2)
		{
			nop++;
			swap(a[i],a[i-1]);
			i-=2;
		}
		if(nop==k)
		{
			for(ll i=1;i<=n;i++)
				cout<<a[i]<<" ";
			cout<<"\n";
		}
		else
			cout<<"-1\n";
	}	
}

int main()
{	
	checkpoint1();
/********************************************************/	
	auto startrrr = high_resolution_clock::now();
	cout << setprecision(20);
/*******************************************************/	
	
	terminAtor();


/*******************************************************/	
	auto stoprrr = high_resolution_clock::now();
	auto duration = duration_cast<microseconds>(stoprrr - startrrr);
	#ifndef ONLINE_JUDGE
		cerr << "Time: " << duration.count()/1000.0<<"\n";
	#endif
/*******************************************************/		
	return 0;
}