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 JAVA):

Solution:
import java.util.*;
import java.lang.*;
public class Codeforces {
    static Scanner sr = new Scanner(System.in);
    static long gcd(long a ,long b) {
        if (b == 0)
            return a;
        return gcd(b, a % b);
    }
    public static void main(String[] args) throws java.lang.Exception {
        StringBuffer ans = new StringBuffer("");
        long T=sr.nextLong();
        while (T-->0) {
            int n=sr.nextInt();
            int k=sr.nextInt();
            if((n%2==0&&k>=n/2)||(n%2==1&&k>n/2))
                ans.append(-1+" "+'\n');
            else
            {
                String s="1 ";
                int i;
                int c=0;
                for(i = 1; i < n&&c<k; i++) {
                   if(i%2==0)
                   {
                       s+=i+" ";
                       ++c;
                   }
                   else
                   {
                       s+=i+2+" ";

                   }
                }
                for(int j=i;j<n;j++)
                    s+=j+1+" ";
                ans.append(s + "" + '\n');
            }
        }
        System.out.println(ans);
    }
}