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 PYTHON3):

Solution:
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
"""
Created on Sun Apr 11 16:46:14 2021

@author: test
"""

t=int(input())

def makeit(k,n):
    L=[i+1 for i in range(n)]
    j=1
    while j<n-1 and k>0:
        L[j]=j+2
        L[j+1]=j+1
        j+=2
        k=k-1
    if k==0:
        return L
    else:
        return -1
for loop in range(t):
    n,k=[int(x) for x in input().split()]
    if makeit(k,n)==-1:
        print(-1)
    else:
        c=makeit(k,n)
        for i in c:
            print(i,end=' ')
        print()