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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Collections;
import java.util.StringTokenizer;

public class A {
 
	public static void main(String[] args) {
		FastScanner fs=new FastScanner();
		int t=fs.nextInt();
		while(t-->0)
		{
			int n = fs.nextInt();
			int k = fs.nextInt();
			ArrayList<Integer> a = new ArrayList<>();
			String ans = "-1";
			int m = (n-1)/2;
			if (k!=0)
			{
			if (k<=m)
			{
				int j = 0;
				int p = 1;
				int q = n;
				boolean is = false;
			for (int i=0; i<n; i++)
			{
				if (!is)
				{
				if (i%2 == 0)
				{
					System.out.print(p+" ");
					p++;
				}
				else
				{
					System.out.print(q+" ");
					q--;
				}
				}
				else
				{
					System.out.print(p+" ");
					p++;
				}
				if (i!=0)
				{
					j++;
				}
				if (j == 2)
				{
					k--;
					j = 0;
					if (k == 0)
						is = true;
				}
			}
			}
			else
				System.out.print(ans);
			}
			else
			{
				for (int i=1; i<=n; i++)
					System.out.print(i+" "); 
			}
			System.out.println(); 
		}
 
	}
 
	static class FastScanner {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
 
		int nextInt() {
			return Integer.parseInt(next());
		}
		int[] readArray(int n) {
			int[] a=new int[n];
			for (int i=0; i<n; i++)
				a[i]=nextInt();
			return a;
		}
		long nextLong() {
			return Long.parseLong(next());
		}
	}
 
}