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 sys
from functools import lru_cache, cmp_to_key
from heapq import merge, heapify, heappop, heappush, nlargest, nsmallest
from math import ceil, floor, gcd, fabs, factorial, fmod, sqrt, inf, log
from collections import defaultdict as dd, deque, Counter as C
from itertools import combinations as comb, permutations as perm
from bisect import bisect_left as bl, bisect_right as br, bisect
from time import perf_counter
from fractions import Fraction
# sys.setrecursionlimit(pow(10, 6))
# sys.stdin = open("input.txt", "r")
# sys.stdout = open("output.txt", "w")
mod = pow(10, 9) + 7
mod2 = 998244353
def data(): return sys.stdin.readline().strip()
def out(*var, end="\n"): sys.stdout.write(" ".join(map(str, var))+end)
def l(): return list(sp())
def sl(): return list(ssp())
def sp(): return map(int, data().split())
def ssp(): return map(str, data().split())
def l1d(n, val=0): return [val for i in range(n)]
def l2d(n, m, val=0): return [l1d(n, val) for j in range(m)]


def dfs(source):
    path.append(source)
    vis[source] = True
    for child in graph[source]:
        if not vis[child]:
            dfs(child)


for _ in range(int(data())):
    n = int(data())
    if n <= 3:
        out(-1)
        continue
    graph = dd(set)
    stack = [1]
    s = set()
    while stack:
        temp = stack.pop()
        for i in range(2, 5):
            if temp + i <= n:
                graph[temp].add(temp+i)
                if temp + i not in s:
                    s.add(temp+i)
                    stack.append(temp+i)
        if temp >= 3:
            for i in range(2, 5):
                if temp-i > 0:
                    graph[temp].add(temp-i)
                    if temp-i not in s:
                        stack.append(temp-i)
                        s.add(temp-i)
    # print(graph)
    for i in range(1, n+1):
        vis = dd(lambda: False)
        path = []
        dfs(i)
        r = True
        for j in range(len(path)-1):
            if 2 <= abs(path[j]-path[j+1]) <= 4:
                continue
            r = False
            break
        if r:
            out(*path)
            break
