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.math.BigInteger;
import java.util.*;
public class Main {
	static Scanner sc=new Scanner(System.in);
	public static void main(String[] args) {
		for(int t=sc.nextInt();t>0;t--) {
			show();
		}
	}
	private static void show() {
		int n=sc.nextInt();
		int k=sc.nextInt();
		int a[]=new int [n+1];
		int arr[]=new int [2*(k+1)+1];
		int sum,minn,maxn;
		for(int i=1;i<=n;i++) {
			a[i]=sc.nextInt();
			if(i>n/2)
	        {
	            sum=a[i]+a[n-i+1];
	            minn=Math.min(a[i],a[n-i+1]);
	            maxn=sum-minn;

	            arr[2]+=2;
	            arr[minn+1]--;
	            arr[maxn+k+1]++;
	            arr[sum]--;
	            arr[sum+1]++;
	        }
		}
		int ans=arr[2];
		///System.out.println(Arrays.toString(arr));
	    for(int i=3;i<=2*k;i++){
	        arr[i]+=arr[i-1];
	        ans=Math.min(arr[i],ans);
	    }
	    
	    System.out.println(ans);
	}
}