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.util.*;
public class pd{
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
		int t = sc.nextInt();
		while(t-->0){
			int n = sc.nextInt();
			int k = sc.nextInt();
			int[] a = new int[n];
			int[] cnt = new int[2*k+1];
			int[] pref = new int[2*k+2];
			for(int i=0;i<n;i++){
				a[i]=sc.nextInt();
			}
			for(int i=0;i<n/2;i++){
				cnt[a[i]+a[n-i-1]]++;
			}
			for(int i=0;i<n/2;i++){
				//int l1=a[i]+1;
				//int r1 = a[n-i+1]+1;
				//int l2 = a[i]+k;
				//int r2 = a[n-i+1]+k;
				int l = Math.min(a[i],a[n-i-1])+1;
				int r = Math.max(a[i],a[n-i-1])+k;
				pref[l]++;
				pref[r+1]--;
			}
			for(int i=1;i<=2*k+1;i++){
				pref[i]+=pref[i-1];
			}
			int ans = Integer.MAX_VALUE;
			for(int sum=2;sum<=2*k;sum++){
				ans = Math.min(ans,pref[sum]-cnt[sum]+(n/2-pref[sum])*2);
			}
			System.out.println(ans);

		}
	}
}