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>
using namespace std;
const int size = 2e5 + 11;
int a[size], b[size];
int main(void) {
  int t, n;
  scanf("%d", &t);
  while (t--) {
    scanf("%d", &n);
    int k = 1;
    if (n % 4 != 0)
      cout << "NO" << endl;
    else {
      int w = n, c = 0, i = 0, j = 0;
      cout << "YES" << endl;
      while (n) {
        if (n % 2 == 0 && n > 2) {
          b[++i] = k++;
          n--;
        }
        if (n % 2 != 0 && n > 2) {
          a[++j] = k++;
          n--;
          c++;
        }
        if (n == 2) {
          a[++j] = w;
          n--;
        } else if (n == 1) {
          b[++i] = w + c;
          n--;
        }
      }
      for (int k = 1; k <= j; k++) printf("%d ", a[k]);
      for (int k = 1; k <= i; k++) printf("%d ", b[k]);
      cout << endl;
    }
  }
}