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.*;

public class Main {
    public static void main(String[] args) throws IOException {
        StreamTokenizer cin = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
        PrintWriter cout = new PrintWriter(new OutputStreamWriter(System.out));
        cin.nextToken();
        int t = (int) cin.nval;

        while (t-- > 0) {
            cin.nextToken();
            int n = (int) cin.nval;
            cin.nextToken();
            int k = (int) cin.nval;
            int[] arr = new int[n];

            for (int i = 0; i < n; i++) {
                cin.nextToken();
                arr[i] = (int) cin.nval;
            }

            //  每对和出现的频率    （不需要更改）
            int[] cnt = new int[2 * k + 1];

            for (int i = 0; i < n / 2; i++) {
                cnt[arr[i] + arr[n - i - 1]]++;
            }
//            cout.print(Arrays.toString(cnt));cout.flush();
            //  pref[x]维护(arr[i],arr[n-i-1])替换一次改成x的次数
            int[] pref = new int[2 * k + 2];
            for (int i = 0; i < n / 2; i++) {
                //  求这个数有多少个区间覆盖了此数
                //  对于每对arr[i]和arr[n-i-1]所能修改的
                //  值域就在Min(arr[i],arr[n-i-1]) + 1记作low
                //  max(arr[i],arr[n-i-1]) + k 记为high   之间
                int low = Math.min(arr[i],arr[n-i-1]) + 1;
                int high = Math.max(arr[i],arr[n-i-1]) + k;

                pref[low]++;
                pref[high+1]--;
            }

            for (int i = 1; i <= 2 * k + 1;i++) {
                pref[i] += pref[i - 1];
            }

            int ans = Integer.MAX_VALUE;

            for (int x = 2; x <= 2 * k; x++) {
                ans = Math.min(ans,(pref[x] - cnt[x]) + (n / 2 - pref[x])* 2);
            }
            cout.println(ans);cout.flush();
        }
    }
}