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 
from sys import stdin, setrecursionlimit, stdout
#setrecursionlimit(1000000)
from collections import deque
from math import sqrt, floor, ceil, log, log2, log10, pi, gcd, sin, cos, asin
def ii(): return int(stdin.readline())
def fi(): return float(stdin.readline())
def mi(): return map(int, stdin.readline().split())
def fmi(): return map(float, stdin.readline().split())
def li(): return list(mi())
def lsi():
    x=list(stdin.readline())
    x.pop()
    return x
def si(): return stdin.readline().rstrip()
#mod=1000000007
res=['YES', 'NO']
############# CODE STARTS HERE #############
x=999//2
a=[0]*1000
a[x], a[x+2], a[x-1], a[x+1]=1, 2, 3, 4
i=x-2
j=x+3
v=5
while i>=0 and j<1000:
    a[i]=v
    a[j]=v+1
    v+=2
    i-=1
    j+=1
#print(a[x])




test_case=ii()
while test_case:
    test_case-=1






    n=ii()
    if n<4:
        print(-1)
    else:
        l, r=498, 502
        if n%2:
            l-=1
            n-=1
        x=(n-4)//2
        l-=x
        r+=x
        print(*a[l:r])
