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>
#pragma GCC optimize("O3")
using namespace std;
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
int32_t main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int64_t t;
  cin >> t;
  while (t--) {
    int64_t n;
    cin >> n;
    if (n % 4)
      cout << "NO"
           << "\n";
    else {
      cout << "YES"
           << "\n";
      int64_t sum = 0;
      for (int64_t i = 2; i <= n; i += 2) cout << i << " ", sum += i;
      for (int64_t i = 1; i < n - 1; i += 2) {
        cout << i << " ", sum -= i;
      }
      cout << sum << "\n";
    }
  }
}