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;
bool sortbysec(const pair<int, int>& a, const pair<int, int>& b) {
  return (a.second < b.second);
}
bool sortbyth(const tuple<long long, long long, long long>& a,
              const tuple<long long, long long, long long>& b) {
  return (get<2>(a) < get<2>(b));
}
struct hash_pair {
  template <class T1, class T2>
  size_t operator()(const pair<T1, T2>& p) const {
    auto hash1 = hash<T1>{}(p.first);
    auto hash2 = hash<T2>{}(p.second);
    return hash1 ^ hash2;
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  unordered_map<pair<int, int>, bool, hash_pair> um;
  long long t;
  cin >> t;
  while (t--) {
    long long n, i;
    cin >> n;
    if ((n / 2) % 2 != 0)
      cout << "NO" << endl;
    else {
      cout << "YES" << endl;
      long long x = 2, sum = 2;
      for (i = 0; i < n / 2; i++) {
        cout << x << " ";
        x += 2;
        sum += x;
      }
      sum -= x;
      x = 1;
      long long z = 1;
      for (i = n / 2; i < n - 1; i++) {
        cout << x << " ";
        x += 2;
        z += x;
      }
      z -= x;
      cout << sum - z << endl;
    }
  }
}