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() {
  long long int t;
  cin >> t;
  while (t--) {
    long long int n;
    cin >> n;
    if (n % 2 != 0) {
      cout << "NO" << endl;
    } else {
      n = n / 2;
      if (n % 2 != 0) {
        cout << "NO" << endl;
      } else {
        cout << "YES" << endl;
        long long int i;
        vector<long long int> v;
        long long int p = 2;
        for (i = 0; i < n; i++) {
          cout << p << " ";
          v.push_back(p);
          p = p + 4;
        }
        int f = 0;
        for (i = 0; i < n; i++) {
          if (f == 0) {
            cout << v[i] - 1 << " ";
            f = 1;
          } else {
            cout << v[i] + 1 << " ";
            f = 0;
          }
        }
        cout << endl;
      }
    }
  }
}