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.io.*;
import java.util.StringTokenizer;

public class A1513 {

    public static void main(String[] args) throws IOException {
        FastReader s = new FastReader();
        BufferedWriter output = new BufferedWriter(
                new OutputStreamWriter(System.out));
      int t,n,k,i=0,j=1;
      t = s.nextInt();
      while (t>0)
      {
          i=0;j=1;
          n = s.nextInt();
          k = s.nextInt();
          if(n==1 && k==0)
              output.write("1");
          else if(n==2 && k==0)
              output.write("1 2");
          else if((n-1)/2 < k)
              output.write("-1");
          else{
              n-=(2*k+1);
              output.write("1");
              while(k>0)
              {
                  i+=2;
                  j+=2;
                  output.write(" "+j+" "+i);
                  k--;
              }
              i++;
              while(n>0)
              {
                  i++;
                  output.write(" "+i);
                  n--;
              }
          }
          output.write("\n");
          output.flush();
          t--;
      }
    }
    
    static 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();
        }

        public int nextInt() { return Integer.parseInt(next()); }

        public long nextLong() { return Long.parseLong(next()); }

        public double nextDouble()
        {
            return Double.parseDouble(next());
        }

        public String nextLine()
        {
            String str = "";
            try {
                str = br.readLine();
            }
            catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }
}