human
Write me a Python code to 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 given the following:
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
assistant
#Here is Python code to 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.

#!/usr/bin/env python3
# -*- coding: utf-8 -*-
"""
Created on Fri May 15 17:09:25 2020

@author: yuandali
"""
class Solution:    
    def getPermutation(self, num):
        self.ans = []
        val = list(range(1, num + 1))
        n = len(val)
        
        def permute(idx):
            if idx == n and len(self.ans) == num: return True
            ans = False
            for i in range(idx, n):    
                if not self.ans or 2<= abs(val[i] - self.ans[-1]) <= 4:
                    self.ans.append(val[i])
                    val[i], val[idx] = val[idx], val[i]
                    ans = ans or permute(idx + 1)
                    val[i], val[idx] = val[idx], val[i]
                    if ans: return True
                    self.ans.pop()
            return ans
        
        permute(0)
        
        
        return self.ans
    
    
#print(Solution().getPermutation(10))

t = int(input())
for _ in range(t):
    ans = Solution().getPermutation(int(input()))
    if not ans: ans = [-1]
    for val in ans:
        print(val, end = " ")
    print()