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 CPP):

Solution:
#include <bits/stdc++.h>
int main() {
  int t, total, i, j, k;
  scanf("%d", &t);
  int len_arr[t];
  for (i = 0; i < t; i++) scanf("%d", &len_arr[i]);
  for (i = 0; i < t; i++) {
    if (len_arr[i] / 2 % 2 != 0)
      printf("NO\n");
    else {
      printf("YES\n");
      total = 0;
      for (j = 0; j < len_arr[i] / 2; j++) {
        total += 2 * (j + 1);
        printf("%d ", 2 * (j + 1));
      }
      for (j = len_arr[i] / 2, k = 1; j < len_arr[i] - 1; j++, k += 2) {
        total -= k;
        printf("%d ", k);
      }
      printf("%d\n", total);
    }
  }
  return (0);
}