A permutation *p* of size *n* is the sequence *p*<sub class="lower-index">1</sub>, *p*<sub class="lower-index">2</sub>, ..., *p*<sub class="lower-index">*n*</sub>, consisting of *n* distinct integers, each of them is from 1 to *n* (1 ≤ *p*<sub class="lower-index">*i*</sub> ≤ *n*).

A lucky permutation is such permutation *p*, that any integer *i* (1 ≤ *i* ≤ *n*) meets this condition *p*<sub class="lower-index">*p*<sub class="lower-index">*i*</sub></sub> = *n* - *i* + 1.

You have integer *n*. Find some lucky permutation *p* of size *n*.
The first line contains integer *n* (1 ≤ *n* ≤ 10<sup class="upper-index">5</sup>) — the required permutation size.
import sys
input = sys.stdin.buffer.readline
# input = sys.stdin.readline
print = sys.stdout.write

# from heapq import heapify,heappush,heappop

def solve():
    n = int(input())
    if(n == 1):
        print("1\n")
        return
    if(n & 1 and (n - 1) % 4 != 0):
        print("-1\n")
        return
    if(n % 2 == 0 and n % 4 != 0):
        print("-1\n")
        return
    ans = [-1 for i in range(n+1)]
    for i in range(1,n+1):
        if(ans[i] == -1):
            req = i + 1
            ans[i] = req
            x = ans[i]
            req = n - i + 1
            while 1:
                if(ans[x] == -1):
                    ans[x] = req
                    req = n - x + 1
                    x = ans[x]
                else:
                    break
    if(n & 1):
        ans[(n + 1) // 2] = (n + 1) // 2
    print(' '.join(map(str,ans[1:])) + "\n")

for _ in range(int(input())):
    solve()

# code by Kavorka
# code by Kavorka
# code by Kavorka
# code by Kavorka
# code by Kavorka
 	   		 	      				 			 	 	 	 	