A permutation of length n is an array p=[p_1,p_2,...,p_n], which contains every integer from 1 to n (inclusive) and, moreover, each number appears exactly once. For example, p=[3,1,4,2,5] is a permutation of length 5.

For a given number n (n ≥ 2), find a permutation p in which absolute difference (that is, the absolute value of difference) of any two neighboring (adjacent) elements is between 2 and 4, inclusive. Formally, find such permutation p that 2 ≤ |p_i - p_{i+1}| ≤ 4 for each i (1 ≤ i < n).

Print any such permutation for the given integer n or determine that it does not exist.

Input

The first line contains an integer t (1 ≤ t ≤ 100) — the number of test cases in the input. Then t test cases follow.

Each test case is described by a single line containing an integer n (2 ≤ n ≤ 1000).

Output

Print t lines. Print a permutation that meets the given requirements. If there are several such permutations, then print any of them. If no such permutation exists, print -1.

Example

Input


6
10
2
4
6
7
13


Output


9 6 10 8 4 7 3 1 5 2 
-1
3 1 4 2 
5 3 6 2 4 1 
5 1 3 6 2 4 7 
13 9 7 11 8 4 1 3 5 2 6 10 12 
# May the Speedforce be with us...
'''
for _ in range(int(input())):
arr=list(map(int,input().split()))
n,k=map(int,input().split())
n=int(input())
s=input()

from collections import defaultdict
d=defaultdict()

d=dict()
s=set()
s.intersection()
s.union()
s.difference()


problem statement achhe se padhna hai
age se bhi dekhna hai, piche se bhi dekhna hai
'''
from math import gcd,ceil
from collections import defaultdict as dd
def lcm(a,b):
	return (a*b)//gcd(a,b)
def inin():
	return int(input())
def inar():
	return list(map(int,input().split()))
def ar(element,size):
	return [element for i in range(size)]
def digitsum(num):
	su=0
	while(num):
		su+=num%10
		num//=10
	return su
def pred(n,k,pointer):
	return (pointer-k)>=n

odd=[1,3,5]
even=[4,2,6]
for i in range(7,1001):
	if i%2:
		odd.append(i)
	else:
		even.append(i)


for _ in range(inin()):
	n=inin()
	if n<4:
		print(-1)
	else:
		arr=odd[0:(n+1)//2]
		arr.reverse()
		arr+=even[0:n//2]
		print(*arr)
