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:
/*   --------------------
    |       LOSER        |
    |     ~NOOBOSS~      |
     --------------------
*/
#include <bits/stdc++.h>
using namespace std;
#define      mxx                 1e18
#define      mnn                 -1e18
#define      Y()                 cout<< "YES" <<endl
#define      N()                 cout << "NO"<<endl
#define      endl                "\n"
#define      Ceil(x,y)           ((x+y-1)/y)
#define      sz(s)               (int)s.size()
#define      angle(x)            double(x * acos(-1) / 180.0)
#define      max_3(a,b,c)        max(a, max(b,c))
#define      min_3(a,b,c)        min(a, min(b,c))
#define      gcd(a,b)            __gcd(a,b)
#define      lcm(a,b)           (a*b)/gcd(a,b)
#define      loser               return 0
#define      ll                  long long
#define      PI                  acos(-1)
#define      mem(a,v)            memset(a,v,sizeof(a))
#define      SORT(v)             sort(v.begin(),v.end())
#define      REV(v)              reverse(v.begin(),v.end())
#define      B                   begin()
#define      E                   end()
#define      V                   vector
#define      F                   first
#define      S                   second
#define      PSB                 push_back
#define      MP                  make_pair
#define      flash               cout.flush()
#define      InTheNameOfGod      ios::sync_with_stdio(0);cin.tie(0); cout.tie(0);
constexpr ll MOD = 998244353;
constexpr ll mod = 1e9  + 7;
int dx[] = {0,0,1,-1};
int dy[] = {1,-1,0,0};
/*-----*/
#define bug1(a)  cerr<<a<<endl;
#define bug2(a,b)  cerr<<a<<" "<<b<<endl;
#define bug3(a,b,c)  cerr<<a<<" "<<b<<" "<<c<<endl;
/*----*/
const ll N=3e6+5;
vector<ll> adj[N];
ll power(ll n,ll p){if(p==0) return 1;if(p==1)return n;if(p%2)return power(n,p-1)*n;else{ll x=power(n,p/2);return x*x;}}
ll modpow(ll a,ll b,ll m){ll ans=1;while(b){if(b&1)ans=(ans*a)%m;b/=2;a=(a*a)%m;}return ans;}
ll nsum(ll num){return (num*(num+1))/2;}
void edge (ll u,ll v) {adj[u].PSB(v) ;adj[v].PSB(u);}
/*------------------START---------------------*/

/*-----*/
void solve(){
   ll n,k;
   string s;
   cin>>n>>k;
   if(n<3){
     if(k) cout<<-1<<endl;
     else{
        for(ll i=1;i<=n;i++) cout<<i<<' ';
        cout<<endl;
     }
     return;
   }

   ll mx=1; ll bal=n-3;
   mx+=bal/2;
   if(k>mx){
    cout<<-1<<endl;
   }
   else{
      cout<<1<<' ';
      ll ct=2;
      for(ll i=2;i<=n;i++){
        if(k){
            cout<<i+1<<' '<<i<<' ';
            --k; i++;
        }
        else{
            cout<<i<<' ';
        }
      }
      cout<<endl;
   }

   //cout << fixed << setprecision(10);
}
/*-----*/
int main(){
   InTheNameOfGod

   ll Test=1;
   cin>>Test;
   while(Test--){
     solve();
    }
  loser;
}
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