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>
#define M_PI       3.14159265358979323846
#define Speed_UP  ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
#define   pb  push_back 
#define   ff  first 
#define   ss  second 
#define sz(x) (int)x.size()
#define all(x) (x).begin(), (x).end()
#warning Remember to change t
inline void setIO();
using namespace std;
typedef long long ll;
typedef long double ld;

int solve(){
  ll n,k;
  cin >> n >> k;

  ll maxPos = (n-1LL)/2;
  if(k>maxPos){
    cout<<"-1\n";
  }else{
    vector<ll>ans(n);
    ll c=0;
    for(int i=0;i<n;i++){
      ans[i]=i+1;
    }
    ll i = n-1;;
    
    while(c!=k){
      if(i-1>=0){
        swap(ans[i],ans[i-1]);
        i=i-2;
        c++;
      }
    }

    for(auto u:ans)
      cout<<u<<" ";

    cout<<"\n";
  }
  
  return 0;
  }
  
int main(){
  Speed_UP
  
  #ifndef ONLINE_JUDGE
  freopen("input.txt", "r", stdin);
  #endif

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


inline void setIO(string name="") {
  #ifndef ONLINE_JUDGE 
  freopen((name+".in").c_str(), "r", stdin); 
  freopen((name+".out").c_str(), "w", stdout);
  #endif
  }