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

public class D{

	public static void main(String[] args) throws Exception{
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		
		int t = Integer.parseInt(br.readLine());

		for(int numCases = 1; numCases <= t; numCases++){
			StringTokenizer st = new StringTokenizer(br.readLine());
			int n = Integer.parseInt(st.nextToken());
			int k = Integer.parseInt(st.nextToken());
			HashMap<Integer, Integer> sums = new HashMap<Integer, Integer>();
			st = new StringTokenizer(br.readLine());
			int[] a = new int[n];

			for(int i = 0; i < n; i++)
				a[i] = Integer.parseInt(st.nextToken());

			FenwickTree maxes = new FenwickTree(2*k+1);
			FenwickTree mins = new FenwickTree(2*k+1);

			for(int i = 0; i < n / 2; i++){
				int sum = a[i] + a[n - 1 - i];
				maxes.adjust(Math.max(a[i], a[n - 1 - i]), 1);	
				mins.adjust(Math.min(a[i], a[n - 1 - i]), 1);	
				if(sums.containsKey(sum))
					sums.put(sum, sums.get(sum) + 1);
				else
					sums.put(sum, 1);
			}

			int min = n / 2;

			for(int i = 0; i < n / 2; i++){
				int sum = a[i] + a[n - 1 - i];
				int total = n / 2 - sums.get(sum);

				total += maxes.rsq(1, sum - k - 1);
				total += mins.rsq(sum, 2*k);
				min = Math.min(min, total);
				// System.out.println("i: " + i);
				// System.out.println("total: " + total);
			}

			System.out.println(min);

		}
		
		br.close();
	}

	static class FenwickTree{

		private ArrayList<Integer> ft = new ArrayList<Integer>();

		public FenwickTree(int n){
			for(int i = 0; i <= n; i++)
				ft.add(0);
		}

	//least significant ones
		public int lso(int i){
			return i & (-1*i);
		}

		public int rsq(int b){
			int sum = 0;
			for(int i = b; i > 0 ; i -= lso(i))
				sum += ft.get(i);
			return sum;
		}

		public int rsq(int a, int b){
			return rsq(b) - (a == 1 ? 0 : rsq(a - 1));
		}

		public void adjust(int k, int v){
			for(int i = k; i < ft.size(); i+= lso(i))
				ft.set(i, ft.get(i) + v);
		}

	}

}