Problem: You are given a positive integer n, it is guaranteed that n is even (i.e. divisible by 2).

You want to construct the array a of length n such that: 

  * The first n/2 elements of a are even (divisible by 2); 
  * the second n/2 elements of a are odd (not divisible by 2); 
  * all elements of a are distinct and positive; 
  * the sum of the first half equals to the sum of the second half (∑_{i=1}^{n/2} a_i = ∑_{i=n/2 + 1}^{n} a_i). 



If there are multiple answers, you can print any. It is not guaranteed that the answer exists.

You have to answer t independent test cases.

Input

The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.

The only line of the test case contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the length of the array. It is guaranteed that that n is even (i.e. divisible by 2).

It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).

Output

For each test case, print the answer — "NO" (without quotes), if there is no suitable answer for the given test case or "YES" in the first line and any suitable array a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) satisfying conditions from the problem statement on the second line.

Example

Input


5
2
4
6
8
10


Output


NO
YES
2 4 1 5
NO
YES
2 4 6 8 1 3 5 11
NO

Solution (in JAVA):

Solution:
//package Contest_636;

import java.io.*;
import java.util.*;
public class D {

	public static void main(String[] args) throws Exception
	{
		Scanner scan = new Scanner(System.in);
		int t = Integer.parseInt(scan.nextLine());
		
		for(int i = 0; i < t; i++)
		{
			int n = scan.nextInt();
			int k = scan.nextInt();
			//scan.nextLine();
			scan.nextLine();
			HashMap<Integer, Integer> sums = new HashMap<Integer, Integer>();
			int[] pref = new int[2*k+2];
			int[] input = new int[n];
			for(int j = 0; j < n; j++)
			{
				input[j] = scan.nextInt();
			}
			
			scan.nextLine();
			for(int j = 0; j < n/2; j++)
			{
				int x = Math.min(input[j], input[n-j-1]);
				int y = Math.max(input[j], input[n-j-1]);
				sums.put((x+y), sums.getOrDefault(x+y, 0)+1);
				pref[x+1] += 1;
				pref[y+k+1] -= 1;
			}
			
			for(int j = 1; j < pref.length; j++)
			{
				pref[j] = pref[j] + pref[j-1];
			}
			
			int min = Integer.MAX_VALUE;
			for(int j = 1; j < pref.length; j++)
			{
				min = Math.min(min, (pref[j]-sums.getOrDefault(j, 0)) + (n/2 - pref[j])*2);
			}
			
			System.out.println(min);
		}
	}

}