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 
import java.util.Scanner;
public class Codeforces {

	public static void main(String[] args) {
		Scanner s=new Scanner(System.in);
			int test=s.nextInt();
		int c[]=new int[test];
		for(int i=0;i<test;i++)
{c[i]=s.nextInt();		
                    }
		for(int j=0;j<test;j++)
		{
			for(int i=0;i<possible(c[j]).length;i++)
			{
				System.out.print(possible(c[j])[i]+" ");
			}System.out.println();
			
		}
		
		
		
		
	}
	static int[] possible(int c)
	{int a[]=new int[c];int t=c;
	int d[]= {-1};
	if (c<=3)
		return d;
	for(int i=0;i<c/2;i++)
	{if (c%2==0)
		{a[i]=t;
	t=t-2;}
	else {
		a[i]=t-1;
		t=t-2;
	}}a[c/2-1]=4;
		a[c/2-2]=2;t=1;
		for(int i=c/2;i<c;i++)
		{
			a[i]=t;
			t=t+2;
			
		}return a;
		
	}
		
		
		
		
		
		
		}
			
		
		
		