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;
inline int read() {
  int ans = 0, mark = 1;
  int t = getchar();
  while (t < '0' || t > '9') {
    if (t == '-') mark = -1;
    t = getchar();
  }
  while (t <= '9' && t >= '0') {
    ans = (ans << 3) + (ans << 1) + t - '0';
    t = getchar();
  }
  return ans * mark;
}
const int INF = 0x3f3f3f3f;
const int N = 1e6 + 5, p = 1e9 + 7;
void init() { return; }
void solve() {
  int(n) = read();
  n /= 2;
  if (n & 1)
    cout << "NO" << endl;
  else
    cout << "YES" << endl;
  if (!(n & 1)) {
    for (int i = 1; i <= n; i++) cout << 2 * i << ' ';
    for (int i = 1; i <= n / 2; i++) cout << 2 * i - 1 << ' ';
    for (int i = 1; i <= n / 2; i++) cout << 2 * (i + n / 2) + 1 << ' ';
    cout << endl;
  }
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  init();
  int(Loop_var) = read();
  while (Loop_var--) solve();
  return 0;
}