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;
int main()
{
  int T;
  cin >> T;
  while (T--)
  {
    int n, k, count = 1;
    cin >> n >> k;
    int A[n];
    memset(A, 0, sizeof(A));
    int l = k;
    for (int i = 0; i < n; i++)
    {
      A[i] = count++;
      if (k > 0)
      {
        k--;
        i++;
      }
    }
    for (int i = 0; i < n; i++)
    {
      if (A[i] == 0)
        A[i] = count++;
    }
    int ans = 0;
    for (int i = 1; i < n - 1; i++)
    {
      if (A[i] > A[i - 1] && A[i] > A[i + 1])
        ans++;
    }
    if (ans == l)
    {
      for (int i = 0; i < n; i++)
        cout << A[i] << " ";
    }
    else
      cout << "-1";
    cout << "\n";
  }
  return 0;
}