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;

public class Solution {

    public static void main(String[] args) throws IOException {
//long startTime = System.currentTimeMillis();
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int numberOfTestCases = Integer.parseInt(br.readLine());

        for (int testCaseNumber = 1; testCaseNumber <= numberOfTestCases; testCaseNumber++) {

            /*
                5 2
             */
            String[] splitLine;
            splitLine = br.readLine().split(" ");
            int n = Integer.parseInt( splitLine[0] );
            int k = Integer.parseInt( splitLine[1] );

            process( n, k );

        }
//System.out.println("C99: " + (System.currentTimeMillis() - startTime));
        br.close();
    }

            
    private static void process( int n, int k ) {

        if( k > 0 && ( n + 1 ) / 2 <= k ) {
            System.out.println( "-1" );
            return;
        }
        
        int i = 1;
        int j = n;
        StringBuilder sb = new StringBuilder();
        
        while( k > 0 ) {
            sb.append( i + " " );
            sb.append( j + " " );
            i++;
            j--;
            k--;
        }
        
        while( j >= i ) {
            sb.append( j + " " );
            j--;
        }
        

        System.out.println( sb.toString().trim() );
    }
}