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 pulse.generics;

import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int T = in.nextInt();
        StringBuilder sb = new StringBuilder();
        for(int t = 0; t < T; ++t) {
            int n = in.nextInt();
            int k = in.nextInt();
            int[] arr = new int[n];
            for(int i = 0; i < n; ++i) arr[i] = in.nextInt();
            Map<Integer, Integer> exact = new HashMap<>();
            int[] prefix = new int[2 * k + 2];
            for(int i = 0; i < n / 2; ++i) {
                int sum = arr[i] + arr[n - 1 - i];
                if(exact.containsKey(sum)) {
                    exact.put(sum, exact.get(sum) + 1);
                } else {
                    exact.put(sum, 1);
                }
                int min = Math.min(arr[i], arr[n - i - 1]) + 1;
                int max = Math.max(arr[i], arr[n - i - 1]) + k;
                prefix[min] += 1;
                prefix[max + 1] -= 1;
            }
            for(int i = 1; i < 2 * k + 2; ++i) {
                prefix[i] += prefix[i - 1];
            }
            int answer = n;
            for(int sum = 2; sum <= 2 * k; ++sum) {
                int pairsWithOneChange = prefix[sum];
                int pairsWithTwoChange = n / 2 - prefix[sum];
                int pairsWithExactSum = exact.get(sum) == null ? 0 : exact.get(sum);
                answer = Math.min(answer, pairsWithOneChange - pairsWithExactSum + 2 * pairsWithTwoChange);
            }
            sb.append(answer + "\n");
        }
        System.out.print(sb.toString());
    }
}