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 
import math
import sys
from collections import defaultdict,Counter,deque,OrderedDict
import bisect
#sys.setrecursionlimit(1000000)
input = iter(sys.stdin.buffer.read().decode().splitlines()).__next__
ilele = lambda: map(int,input().split())
alele = lambda: list(map(int, input().split()))
def list2d(a, b, c): return [[c] * b for i in range(a)]
#def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
#INF = 10 ** 18
#MOD = 1000000000 + 7
from itertools import accumulate,groupby
import sys
A=[];B = []
for i in range(1,1001):
    if i%2 == 0:
        A.append(i)
    else:
        B.append(i)
A[0],A[1] = A[1],A[0]
        
#print(A,B)
for _ in range(int(input())):
    N = int(input())
    if N<4:
        print(-1)
    else:
        x = A[:(N//2)]
        y = B[:math.ceil(N/2)]
        print(*(y[::-1] + x))
    
    