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.BufferedWriter;
import java.io.IOException;
import java.io.OutputStreamWriter;
import java.util.Arrays;
import java.util.Scanner;

public class d {
    public static void main(String[] args) throws IOException {
        Scanner std = new Scanner(System.in);
        BufferedWriter log = new BufferedWriter(new OutputStreamWriter(System.out));
	    StringBuilder strBu = new StringBuilder();

        int t = std.nextInt();
        while (t-- > 0) {
            int n = std.nextInt();
            int k = std.nextInt();
            int[] vals = new int[n];
            int[] noChange = new int[k*2 + 4];
            int[] oneChange = new int[k*2 + 4];
            for (int i = 0; i < n; i++) {
                vals[i] = std.nextInt();
            }
            for (int i = 0; i < n/2; i++) {
                int v1 = vals[i];
                int v2 = vals[n - 1 - i];

                int total = v1 + v2;
                noChange[total]++;

                int minVal = Math.min(v1, v2) + 1;
                int maxVal = Math.max(v1, v2) + k + 1;
                oneChange[minVal]++;
                oneChange[maxVal]--;
            }
            long prefixSum = 0;
            long ans = Integer.MAX_VALUE;
//            System.out.println(Arrays.toString(noChange));
//            System.out.println(Arrays.toString(oneChange));
            for (int i = 2; i <= 2*k; i++) {
                //i is the x val we will try
                long pairsNoChange = noChange[i]; //# of pairs that are perfect for i
                prefixSum += oneChange[i]; //from 2...i sum
                long pairsOneChange = prefixSum - pairsNoChange; //# of pairs that need 1 value changed for i
                long pairsTwoChange = n/2 - (pairsNoChange + pairsOneChange);
                long iAns = pairsOneChange + pairsTwoChange*2;
                ans = Math.min(ans, iAns);
            }
            log.write(ans + "\n");
        }
        log.flush();
	

    }
}

/**
Remember to check for overflow
Does number need to be greater than a lower bound?
**/