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.BufferedReader;
import java.io.InputStreamReader;
import java.util.*;
public class CF_636_D {

	public static void main(String[] args) {

        Scanner in = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
        int t = in.nextInt(); // Scanner has functions to read ints, longs, strings, chars, etc.
        for (int i = 1; i <= t; ++i) {
         int n = in.nextInt();
         int k = in.nextInt();
         int arr[] = new int[n];
         HashMap<Integer,Integer> hash = new HashMap<Integer, Integer>();
//         int range[][] = new int[n/2][2];
         int[] pre = new int[2*k +2];
         for(int j =0; j <n; j++) {
        	 arr[j] = in.nextInt();
         }
         for(int j =0; j < n/2; j++) {
        	 if(hash.containsKey(arr[j]+ arr[n-j-1])) {
        		 int temp = hash.get(arr[j]+ arr[n-j-1]);
        		 hash.put(arr[j] + arr[n-j-1], ++temp);
        	 }else {
        		 hash.put(arr[j] + arr[n-j-1], 1);  }
        	 
        	 int min = Math.min(arr[j], arr[n-j-1]);
//        	 range[j][0] = min+1;
        	 int max= arr[j]+ arr[n-j-1]- min;
        	 pre[min+1]++;
        	 pre[max+1+k]--;
         }
         
         for(int j = 3; j < 2*k +1; j++) {
        	 pre[j] = pre[j-1] + pre[j];
         }
         int min = Integer.MAX_VALUE;
         for(int k1 = 2; k1 <= 2*k; k1++) {
        	 int a= 0; //number of pairs that require zero change to get k1
        	 if(hash.containsKey(k1)) {
        		 a = hash.get(k1);}
        	 int b = pre[k1] - a; //number of pairs that require 1 change to get k1 

        	 
//        	 #############causing TLE
//        		 for(int j =0; j < n/2; j++) {
//            		 if(k1 >= range[j][0] && k1 <= range[j][1]) {
//            			 if(k1 != range[j][0] -1 + range[j][1] - k) {
//            				b++; 
//            			 }
//            			 
//            		 }
//            	 }
        	 
        	 
        	 int c = n/2 -a - b; //number of pairs needing 2 changes 
        	 min = Math.min(min, b+ 2*c);	 
         }
         System.out.println(min);
        
        }
      }

	
}