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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  int i, t;
  cin >> t;
  for (i = 0; i < t; i++) {
    int n;
    cin >> n;
    if (n % 4 != 0) {
      cout << "NO" << endl;
    } else {
      cout << "YES" << endl;
      vector<int> a(n);
      int k = 2, l = 1, sum = 0, sumo = 0;
      for (int j = 0; j < n / 2; j++) {
        a[j] = k;
        k += 2;
        sum += a[j];
        if (j + n / 2 != n - 1) {
          a[j + n / 2] = l;
          l += 2;
          sumo += a[j + n / 2];
        } else {
          a[j + n / 2] = sum - sumo;
        }
      }
      for (int j = 0; j < n; j++) cout << a[j] << " ";
      cout << endl;
    }
  }
  return 0;
}