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 CF1343D extends PrintWriter {
	CF1343D() { super(System.out); }
	Scanner sc = new Scanner(System.in);
	public static void main(String[] $) {
		CF1343D o = new CF1343D(); o.main(); o.flush();
	}

	void main() {
		int t = sc.nextInt();
		while (t-- > 0) {
			int n = sc.nextInt();
			int k = sc.nextInt();
			int[] aa = new int[n];
			for (int i = 0; i < n; i++)
				aa[i] = sc.nextInt();
			int[] dd = new int[k + k + 1 + 1];
			for (int i = 0, j = n - 1; i < j; i++, j--) {
				int a = aa[i], b = aa[j];
				int l, r;
				l = a + b + 1; r = k + Math.max(a, b);
				if (l <= r) {
					dd[l]++;
					dd[r + 1]--;
				}
				l = r + 1; r = k + k;
				if (l <= r) {
					dd[l] += 2;
					dd[r + 1] -= 2;
				}
				l = 1 + Math.min(a, b); r = a + b - 1;
				if (l <= r) {
					dd[l]++;
					dd[r + 1]--;
				}
				r = l - 1; l = 1 + 1;
				if (l <= r) {
					dd[l] += 2;
					dd[r + 1] -= 2;
				}
			}
			int ans = n;
			for (int a = 1 + 1; a <= k + k; a++)
				ans = Math.min(ans, dd[a] += dd[a - 1]);
			println(ans);
		}
	}
}