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 
///package specialpermutation;

import java.util.*;
import java.io.*;

public class specialpermutation {
	public static void main(String[] args) throws IOException {
		BufferedReader fin = new BufferedReader(new InputStreamReader(System.in));
		int t = Integer.parseInt(fin.readLine());
		StringBuilder fout = new StringBuilder();
		while(t-- > 0) {
			int n = Integer.parseInt(fin.readLine());
			if(n <= 3) {
				fout.append("-1\n");
			}
			else if(n == 4) {
				fout.append("3 1 4 2\n");
			}
			else if(n == 5) {
				fout.append("1 3 5 2 4\n");
			}
			else {
				int pointer = 1;
				int taken = 0;
				for(int i = 0; i < n; i++) {
					fout.append(pointer).append(" ");
					if(pointer % 2 == 1 && pointer + 2 <= n) {
						pointer += 2;
					}
					else if(pointer % 2 == 1 && pointer + 2 > n) {
						if(n % 2 == 0) {
							pointer -= 3;
							fout.append(pointer).append(" ");
							taken = pointer;
							pointer += 4;
							fout.append(pointer).append(" ");
							pointer -= 2;
							i += 2;
						}
						else {
							pointer -= 3;
							fout.append(pointer).append(" ");
							taken = pointer;
							pointer += 2;
							fout.append(pointer).append(" ");
							pointer -= 4;
							i += 2;
						}
					}
					else if(pointer % 2 == 0) {
						//System.out.println("pointer is even: " + pointer + " " + i);
						pointer -= 2;
						if(pointer == taken) {
							pointer -= 2;
						}
					}
				}
				fout.append("\n");
			}
		}
		System.out.print(fout);
	}
}
