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(NULL);
  int t;
  cin >> t;
  while (t--) {
    long long int n;
    cin >> n;
    long long int x = n / 2;
    if (x % 2 != 0) {
      cout << "NO" << endl;
    } else {
      cout << "YES" << endl;
      long long a[n / 2];
      long long sum = 0;
      for (long long i = 0; i < n / 2; i++) {
        a[i] = 2 * i + 2;
        sum += a[i];
      }
      long long b[n / 2];
      long long sum2 = 0;
      for (long long i = 0; i < n / 2; i++) {
        b[i] = 2 * i + 1;
        sum2 += b[i];
      }
      if (sum == sum2) {
        for (int i = 0; i < n / 2; i++) {
          cout << a[i] << " ";
        }
        for (int i = 0; i < n / 2; i++) {
          cout << b[i] << " ";
        }
      } else {
        long long c = 0;
        while (sum != sum2) {
          sum2 += 2;
          c++;
        }
        b[(n / 2) - 1] += 2 * c;
        for (int i = 0; i < n / 2; i++) {
          cout << a[i] << " ";
        }
        for (int i = 0; i < n / 2; i++) {
          cout << b[i] << " ";
        }
      }
      cout << endl;
    }
  }
}