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.util.*;
import java.awt.*;
public class D
{
    BufferedReader in;
    PrintWriter ob;
    StringTokenizer st;
 
    private int max = 400010;
  
    public static void main(String[] args) throws IOException {
        new D().run();
    }
    void run() throws IOException {
        in=new BufferedReader(new InputStreamReader(System.in));
        ob=new PrintWriter(System.out);
        int t = ni();
        while(t-->0)
            solve();
        ob.flush();
    }
    void solve() throws IOException {
        int n = ni();
        int k = ni();
        int a[] = nia(0 , n);
    
        HashMap<Integer , Integer> hm = new HashMap<>(); 
        int dp[] = new int[2*k+12];
        for(int i=0 ; i<n/2 ; i++) {
            int sum=a[i]+a[n-i-1];
            hm.put( sum , hm.getOrDefault( sum , 0 ) + 1 );
 
            int x1 = sum - Math.max( a[i]-1 , a[n-i-1]-1 );
            int x2 = sum + Math.max( k - a[i] , k - a[n-i-1] );
 
            dp[x1]+=1;
            dp[sum]-=1;
            dp[sum+1]+=1;
            dp[x2+1]-=1;
        }
 
        for(int i=1 ; i<2*k+10 ; i++)
            dp[i]+=dp[i-1];
 
        int ans = Integer.MAX_VALUE;
        for(int x=2 ; x<=k+k ; x++) {
            int alreadyPresent = hm.getOrDefault(x,0);
            int needOneChange = dp[x];
            int needTwoChanges = n - (needOneChange*2) - (alreadyPresent*2);
            
            ans = Math.min( ans , needOneChange + needTwoChanges );
        }
 
        ob.println(ans);
    }
 
    String ns() throws IOException {
        return nextToken();
    }
    long nl() throws IOException {
        return Long.parseLong(nextToken());
    }
    int ni() throws IOException {
        return Integer.parseInt(nextToken());
    }
    double nd() throws IOException {
        return Double.parseDouble(nextToken());
    } 
    String nextToken() throws IOException {
        if(st==null || !st.hasMoreTokens())
            st=new StringTokenizer(in.readLine());
        return st.nextToken();
    }
    int[] nia(int start,int b) throws IOException {
        int a[]=new int[b];
        for(int i=start;i<b;i++)
            a[i]=ni();
        return a;
    }
    long[] nla(int start,int n) throws IOException {
        long a[]=new long[n];
        for (int i=start; i<n ;i++ ) {
            a[i]=nl();
        }
        return a;
    }
    
    public void tr(Object... o) {
     System.out.println(Arrays.deepToString(o));
    }
}