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 BIG = 1e5 + 9;
const int INF = 1e9 + 9;
const long long BINF = 1e18 + 9;
const double SML = (1e-7);
long long t, n, i, sum1, sum2;
bool check() {
  int num = n / 2;
  if (num % 2 == 0) {
    return true;
  }
  return false;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> t;
  while (t--) {
    cin >> n;
    sum1 = sum2 = 0;
    bool go = check();
    if (!go) {
      cout << "NO\n";
    } else {
      cout << "YES\n";
      vector<int> ans;
      for (int i = 1, j = 2; i <= n / 2; i++, j = j + 2) {
        sum1 = sum1 + j;
        ans.push_back(j);
      }
      for (int i = 1, j = 1; i < n / 2; i++, j = j + 2) {
        sum2 = sum2 + j;
        ans.push_back(j);
      }
      ans.push_back(sum1 - sum2);
      for (i = 0; i < ans.size(); i++) {
        cout << ans[i] << ' ';
      }
      cout << "\n";
    }
  }
  return 0;
}