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 collections import defaultdict as dc
from collections import Counter 
from heapq import *
import math
from bisect import bisect_left,bisect
#bisect gives x and p[x] is element greater than it and out of bound for last one 
#p[x-1] gives equal or smaller and no error for any element.
import sys
from collections import deque as dq
from heapq import heapify,heappush,heappop
mod=10**9 +7
def sinp():
    p=str(input())
    return p
def inp(): 
    p=int(input())
    return p
def line():
    p=list(map(int,input().split()))
    return p
def read_mat():
    n=inp()
    a=[]
    for i in range(n):
        a.append(line())
    return a
def digit(n):
    s=str(n)
    p=0
    for i in s:
        p+=(int(i))**2
    return p
def check(a):
    n=len(a)
    for i in range(1,n):
        if abs(a[i]-a[i-1])<2 or abs(a[i]-a[i-1])>4:
            return "NO"
    return "YES"
def solve(n):
    p=[0]*n
    if a<4:
        return -1
    if a==4:
        return [3,1,4,2]
    z=n//2
    p[z]=1
    l,r=z,z
    l=l-1
    r=r+1
    p[l]=4
    p[r]=3
    l=l-1
    r=r+1
    p[l]=2
    if n>=5:
        p[r]=5
        x,y=5,4
        l,r=l-1,r+1
        while(l>=0 and r<n):
            p[r]=x+2
            p[l]=y+2
            x=p[r]
            y=p[l]
            l,r=l-1,r+1
        while(l>=0):
            p[l]=y+2
            y=p[l]
            l=l-1
        while(r<n):
            p[r]=x+2
            x=p[r]
            r+=1
    return p

            
    
for test in range(inp()):
    a=inp()
    l=solve(a)
    if l==-1:
        print(-1)
    else:
        print(*l)
    