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.*;
import java.io.*;
public class  a  {
    static final int mod = 1000000007; 
    static FastReader sc = new FastReader();  
    static PrintWriter out = new PrintWriter(System.out);
        
    public static void main (String[] args) throws java.lang.Exception {
        int t = sc.nextInt();
        while (t-->0) {
            int n = sc.nextInt();
            int k = sc.nextInt();
            int arr[] = new int[n];
            int maxPeak = (n-1)/2;
            if (k > maxPeak) {
                out.println(-1);
            }else {
                int i = 1;
                int cnt = 0 ,value = n-k+1;
                while(cnt < k) {
                    arr[i] = value;
                    value++;
                    i +=2;
                    cnt++;
                }
                value = 1;
                for ( i = 0 ; i < n; i++) {
                    if (arr[i] == 0) {
                        arr[i] = value;
                        value++;
                    }
                    out.print(arr[i]+" ");
                }
                out.println();
            }

        }
        out.flush();        
    }
}
class FastReader{ 
     
        BufferedReader br; 
        StringTokenizer st; 
  
        public FastReader() 
        { 
            br = new BufferedReader(new
                     InputStreamReader(System.in)); 
        } 
  
        String next() 
        { 
            while (st == null || !st.hasMoreElements()) 
            { 
                try
                { 
                    st = new StringTokenizer(br.readLine()); 
                } 
                catch (IOException  e) 
                { 
                    e.printStackTrace(); 
                } 
            } 
            return st.nextToken(); 
        } 
  
        int nextInt() 
        { 
            return Integer.parseInt(next()); 
        } 
  
        long nextLong() 
        { 
            return Long.parseLong(next()); 
        } 
  
        double nextDouble() 
        { 
            return Double.parseDouble(next()); 
        } 
  
        String nextLine() 
        { 
            String str = ""; 
            try
            { 
                str = br.readLine(); 
            } 
            catch (IOException e) 
            { 
                e.printStackTrace(); 
            } 
            return str; 
        } 
    }