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.* ; 

public class Mpeaks {
		
		public static void solve(int n,int k) {
			if(k > (n-1)/2) 
				System.out.println(-1); 

			else {
				int max = n ;
				int j = 1; 
				int flag = 1;
				while(j <= n && flag-1 != n ) {

					if(flag%2 == 0 && k>0){
						System.out.print(max+" ");
						k--;
						max--;
                    }

					else{
						System.out.print(j+" ");
						j++;
						} 
                    flag++;

				}
					System.out.println();
			}
		}  


		public static void main(String args[]) {

			Scanner in = new Scanner(System.in); 
			int t=in.nextInt(); 
			while(t > 0) {
			int n = in.nextInt(); 
			int k = in.nextInt(); 
			solve(n,k);
				t--;
			}

		}

}