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 INF = 0x3f3f3f3f, max_n = 200005, mod = 1e9 + 7;
inline long long lowbit(long long x) { return x & (-x); }
inline long long gcd(long long a, long long b) {
  return b == 0 ? a : gcd(b, a % b);
}
inline long long min(long long a, long long b) { return a < b ? a : b; }
inline long long max(long long a, long long b) { return a > b ? a : b; }
double pi = acos(-1);
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(nullptr);
  int t;
  cin >> t;
  long long n;
  while (t--) {
    cin >> n;
    if ((n / 2) % 2 == 1)
      cout << "NO" << '\n';
    else {
      int x = 1;
      long long y = n / 4 * (n + 2) - (n / 2 - 1) * (n - 2) / 2;
      if (y % 2 == 1 && y > n - 3) {
        cout << "YES" << '\n';
        for (int i = 1; i <= n / 2; i++) {
          cout << i * 2 << ' ';
        }
        for (int i = n / 2 + 1; i < n; i++) {
          cout << x << ' ';
          x += 2;
        }
        cout << y << '\n';
      } else
        cout << "NO" << '\n';
    }
  }
  return 0;
}