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

public class Problem_D {
	static final int INF = Integer.MAX_VALUE / 2;

	public static void main(String[] args) {
		InputReader in = new InputReader();
		StringBuilder out = new StringBuilder();

		int T = in.nextInt();
		while (T-- > 0) {
			int N = in.nextInt();
			int K = in.nextInt();
			int[] A = new int[N];
			for (int i = 0; i < N; i++) {
				A[i] = in.nextInt();
			}
			int[] B = new int[2 * K + 2];
			for (int i = 0; i < N / 2; i++) {
				int min = Math.min(A[i], A[N - i - 1]);
				int max = Math.max(A[i], A[N - i - 1]);
				int sum = A[i] + A[N - i - 1];
				B[2] += 2;
				B[min + 1]--;
				B[max + K + 1]++;
				B[sum]--;
				B[sum + 1]++;
			}
			for (int i = 2; i <= 2 * K; i++) {
				B[i] += B[i - 1];
			}
			int ans = INF;
			for (int i = 2; i <= 2 * K; i++) {
				ans = Math.min(ans, B[i]);
			}
			out.append(ans).append('\n');
		}

		System.out.print(out);
	}

	static class InputReader {
		public BufferedReader reader;
		public StringTokenizer st;

		public InputReader() {
			reader = new BufferedReader(new InputStreamReader(System.in));
		}

		public String next() {
			while (st == null || !st.hasMoreTokens()) {
				st = new StringTokenizer(nextLine());
			}
			return st.nextToken();
		}

		public String nextLine() {
			try {
				return reader.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return null;
		}

		public int nextInt() {
			return Integer.parseInt(next());
		}
	}
}