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.HashMap;
import java.util.Scanner;

public class codefcf {
    public static void main(String[] args) {
        Scanner s = new Scanner(System.in);

        int t=s.nextInt();
        while(t-->0) {
            int n=s.nextInt(),k=s.nextInt(),max=0;
            HashMap<Long,Integer> map=new HashMap();
            long a[]=new long[n],pre[]=new long[k*2+10];
            long ans=(long)(n/2);
            for(int i=0;i<n;i++) {
                a[i]=s.nextLong();
            }
            for(int i=0;i<n/2;i++) {
                long ax=Math.max(a[i], a[n-1-i]),az=Math.min(a[i], a[n-1-i]);
                if(!map.containsKey(ax+az))
                    map.put(az+ax, 0);
                map.put(ax+az, map.get(az+ax)+1);
                pre[(int)az+1]+=1;
                pre[(int)ax+k+1]-=1;
            }
            for(int i=1;i<pre.length;i++)
                pre[i]+=pre[i-1];
            for(long o:map.keySet()) {
                int x=(int)o;
                long lol=pre[x]-map.get(o)+2*((n/2)-pre[x]);
                ans=Math.min(ans,lol);
            }
            System.out.println(ans+" ");
        }
    }
}