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;

// getting set element by index using std::advance(it,index-1) and *it
template< class InputIt, class Distance >
void advance( InputIt& it, Distance n );


typedef long long ll;
typedef long double ld;

typedef pair<int, int> pi;
typedef pair<ll,ll> pl;

typedef vector<int> vi;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
#define loop(i,n) for(int i=0;i<n;i++)
#define floop(i,a,n) for(int i=a;i<n;i++)
#define rloop(i,n,a) for(int i=n-1;i>=a;i--)

#define all(x) x.begin(),x.end()
#define ins insert
#define pb push_back
#define mp make_pair
#define F first
#define S second
#define inf INT_MAX
#define ninf INT_MIN
#define M 1000000

const int mod = 1e9 + 7;
const int N = 2e4 + 4;
bool isprime(ll n){
  for(int i=2;i*i<=n;i++){
    if(n%i==0)return false;
  }
  return true;

}
// sorted by second if = then by first greater first
// struct myComp {
//      bool operator()(
//         pair<ll, ll> a,
//         pair<ll, ll> b)

//     {
//         if(a.second==b.second)
//         return a.first<b.first;
//         return a.second < b.second;
//     }


int main()
{

    ios_base::sync_with_stdio(0); cin.tie(0);

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

  }


}