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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
 
public class constantpalin {
    public static void main(String argv[]) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int t = Integer.parseInt(br.readLine());
        while(t-- > 0) {
            String s[] = br.readLine().split(" ");
            String inp[] = br.readLine().split(" ");
            int k = Integer.parseInt(s[1]);
            int n =Integer.parseInt(s[0]);
            int ar[] = new int[n];
            for(int i = 0; i < n; i++) {
                ar[i] = Integer.parseInt(inp[i]);
            }
            int a[] = new int[(2*k)+100], cnt[] = new int[(2*k)+100];
            Arrays.fill(a, 0);Arrays.fill(cnt, 0);
            for(int i = 0; i < n/2; i++) {
                a[ar[i]+ar[n-i-1]]++;
                int mn = Math.min(ar[i], ar[n-i-1]);
                int mx = Math.max(ar[i], ar[n-i-1]);
                mn++;mx+=k+1;
                cnt[mn]++;
                cnt[mx]--;
            }
            int ans = n;
            for(int i = 1; i <= 2*k; i++) {
                cnt[i] += cnt[i-1];
                int cur = n - a[i] - cnt[i];
                ans = Math.min(ans, cur);
            }
 
            
        
            System.out.println(ans);
        }
 
    }
    
    
 
}