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.*;
import java.lang.*;
import java.io.*;
 
public class Codechef
{
   
    static int MAX = (int)1e7;
    static StringBuilder s = new StringBuilder();
	public static void main (String[] args) throws java.lang.Exception
	{
	    InputReader input = new InputReader(System.in);
	    //Scanner input = new Scanner(System.in);
	    OutputWriter out = new OutputWriter(System.out);
	    int test = input.readInt();
	    while(test-->0){
	        int n = input.readInt();
	        int k = input.readInt();
	        int arr[] = new int[n];
	        for(int i =0;i<n;i++)arr[i] = input.readInt();
	        int size = n>>1;
	        int size2 = 2*k;
	        int cache[] = new int[size2+1];
	        for(int i=0;i<size;i++){
	            int a = arr[i]+arr[n-1-i];
	            cache[a]++;
	        }
	        int prefix[] = new int[size2+2];
	        
	        for(int  i =0;i<size;i++){
	            int min =min(arr[i],arr[n-i-1])+1;
	            int max = max(arr[i],arr[n-i-1])+k;
	            prefix[min]++;
	            prefix[max+1]--;
	        }
	        for(int i =1;i<=size2;i++){
	            prefix[i]+=prefix[i-1];
	           // System.out.print(prefix[i]+" ");
	        }
	        int res =MAX;
	        for(int i=1;i<=size2;i++){
	            res = min(res,(prefix[i]-cache[i])+(size-prefix[i])*2);
	        }
	        out.printLine(res);
	       //s.append(res);
	       //s.append("\n");
	        
	    }
	   // out.printLine(s);
 
	}
	public static int min (int a,int b){
	    return a<=b?a:b;
	}
	public static int max(int a,int b){
	    return a>=b?a:b;
	}
 
 
}
 class InputReader{
		private InputStream stream;
		private byte[] buf = new byte[1024];
		private int curChar;
		private int numChars;
		private SpaceCharFilter filter;
		
		public InputReader(InputStream stream) {
			this.stream = stream;
		}
		
		public int read() {
			if (numChars == -1) {
				throw new InputMismatchException();
			}
			if (curChar >= numChars) {
				curChar = 0;
				try {
					numChars = stream.read(buf);
				} catch (IOException e) {
					throw new InputMismatchException();
				}
				if (numChars <= 0) {
					return -1;
				}
			}
			return buf[curChar++];
		}
		
		public int readInt() {
			int c = read();
			while (isSpaceChar(c)) {
				c = read();
			}
			int sgn = 1;
			if (c == '-') {
				sgn = -1;
				c = read();
			}
			int res = 0;
			do {
				if (c < '0' || c > '9') {
					throw new InputMismatchException();
				}
				res *= 10;
				res += c - '0';
				c = read();
			} while (!isSpaceChar(c));
			return res * sgn;
		}
		
		public String readString() {
			int c = read();
			while (isSpaceChar(c)) {
				c = read();
			}
			StringBuilder res = new StringBuilder();
			do {
				res.appendCodePoint(c);
				c = read();
			} while (!isSpaceChar(c));
			return res.toString();
		}
		
		public double readDouble() {
			int c = read();
			while (isSpaceChar(c)) {
				c = read();
			}
			int sgn = 1;
			if (c == '-') {
				sgn = -1;
				c = read();
			}
			double res = 0;
			while (!isSpaceChar(c) && c != '.') {
				if (c == 'e' || c == 'E') {
					return res * Math.pow(10, readInt());
				}
				if (c < '0' || c > '9') {
					throw new InputMismatchException();
				}
				res *= 10;
				res += c - '0';
				c = read();
			}
			if (c == '.') {
				c = read();
				double m = 1;
				while (!isSpaceChar(c)) {
					if (c == 'e' || c == 'E') {
						return res * Math.pow(10, readInt());
					}
					if (c < '0' || c > '9') {
						throw new InputMismatchException();
					}
					m /= 10;
					res += (c - '0') * m;
					c = read();
				}
			}
			return res * sgn;
		}
		
		public long readLong() {
			int c = read();
			while (isSpaceChar(c)) {
				c = read();
			}
			int sgn = 1;
			if (c == '-') {
				sgn = -1;
				c = read();
			}
			long res = 0;
			do {
				if (c < '0' || c > '9') {
					throw new InputMismatchException();
				}
				res *= 10;
				res += c - '0';
				c = read();
			} while (!isSpaceChar(c));
			return res * sgn;
		}
		
		public boolean isSpaceChar(int c) {
			if (filter != null) {
				return filter.isSpaceChar(c);
			}
			return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
		}
		
		public String next() {
			return readString();
		}
		
		public interface SpaceCharFilter {
			public boolean isSpaceChar(int ch);
		}
	}
	
 class OutputWriter {
		private final PrintWriter writer;
		
		public OutputWriter(OutputStream outputStream) {
			writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
		}
		
		public OutputWriter(Writer writer) {
			this.writer = new PrintWriter(writer);
		}
		
		public void print(Object... objects) {
			for (int i = 0; i < objects.length; i++) {
				if (i != 0) {
					writer.print(' ');
				}
				writer.print(objects[i]);
			}
			writer.flush();
		}
		
		public void printLine(Object... objects) {
			print(objects);
			writer.println();
			writer.flush();
		}
		
		public void close() {
			writer.close();
		}
		
		public void flush() {
			writer.flush();
		}
	}
	   
//********************************************************************************************************************************
//////////////////////////////////NOTESSS//////////////////////////////////////////////////////////////////////
/*remember substring length you are dealing with take only those substrings
chain like pattern -> linked list (or) to from Arrays
finding degree can be done during input scanning itself(ok&=true:false flag trick)
longest subsquence can be solved in O(n) itself incase of consecutive numbers why try all subsequence go o dp[i-1]
(*******consecutive numberssssss*********)
initialize all global variables and collection ds
swapping changes the arr so make sure initialize with the original again(careful from where to start after swapping)
missing out character when comparing ==0 instead of =='0'
keep going until a condition and start off from first ****stack***
meet in the middle algo
one dfs call will tell u the dfs of all nodes
2d prefix array for max occurence of number in a range.also storing positions in case of duplicate numbers can
help bring down the complexity to o(N) and considering ranges based on positions
****___****make elements equal -----> wat value of x can it be (or) what can the values range from in that case
use prefix arrays. if range is small then answers is that chech minimum sum possible max sum possible and any 
num in this range could be the answer.
when u want to apply changes for a range PREFIX SUMS
*/