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:
from queue import Queue
from itertools import accumulate
from io import BytesIO, IOBase
from itertools import combinations
import threading
from collections import deque
from functools import reduce, cmp_to_key
import sys
from math import sqrt
import heapq
import random
import math
from collections import defaultdict
from collections import Counter
import io
import os
import bisect
from fractions import Fraction


# sys.setrecursionlimit(200000)
# input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline


def input():
    return sys.stdin.readline().strip()


def iinput():
    return int(input())


def tinput():
    return input().split()


def rinput():
    return map(int, tinput())


def rlinput():
    return list(rinput())


mod = int(1e9)+7


def factors(n):
    return set(reduce(list.__add__,
                      ([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0)))


def swap(arr, a, b):
    arr[a], arr[b] = arr[b], arr[a]
# ----------------------------------------------------
# sys.stdin = open('input.txt', 'r')
# sys.stdout = open('output.txt', 'w')


# ----------------------------------------------------------------


t = iinput()
# t = 1
for _ in range(t):
    n, k = rinput()
    if k == 0:
        print(*[i for i in range(1,n+1)])
        continue
    maxval = int(math.ceil((n-2)/2.0)) 
    if (maxval < k) or (n==2) or n==1:
        print(-1)
        continue
    val = n
    moves = k
    arr = [1]
    temp = 2
    while moves:
        arr.append(val)
        arr.append(temp)
        temp += 1
        val-=1
        moves -= 1
    for i in range(temp, n - k+1):
        arr.append(i)
    print(*arr)