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 long long MOD = 1000000007;
template <class T>
ostream& operator<<(ostream& os, vector<T> v) {
  cout << "[";
  for (auto& i : v) cout << i << ", ";
  cout << "]"
       << "\n";
  return os;
}
template <typename K, typename V>
ostream& operator<<(ostream& os, pair<K, V> p) {
  os << "{" << p.first << ", " << p.second << "} ";
  return os;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    if (n % 4) {
      cout << "NO"
           << "\n";
      continue;
    }
    cout << "YES"
         << "\n";
    int ans = 0;
    for (int i = 1; i <= n / 2; ++i) {
      cout << i * 2 << " ";
      ans += i * 2;
    }
    for (int i = 1; i < n / 2; ++i) {
      cout << i * 2 - 1 << " ";
      ans -= i * 2 - 1;
    }
    cout << ans << "\n";
  }
  return 0;
}