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 maxn = 200009;
const long long MOD = 119 << 23 | 1;
class {
 public:
  int a[maxn];
  void solve() {
    int t;
    cin >> t;
    while (t--) {
      int n, k;
      cin >> n >> k;
      for (int i = 1; i <= n; ++i) cin >> a[i];
      vector<long long> sum(2 * k + 2);
      for (int i = 1; i <= n / 2; ++i) {
        int j = n - i + 1;
        if (a[i] > a[j]) swap(a[i], a[j]);
        sum[2] += 2;
        sum[a[i] + 1] -= 2;
        sum[a[i] + 1]++;
        sum[a[i] + a[j] - 1 + 1]--;
        sum[a[i] + a[j] + 1]++;
        sum[a[j] + k + 1]--;
        sum[a[j] + k + 1] += 2;
        sum[2 * k + 1] -= 2;
      }
      for (int i = 1; i <= 2 * k; ++i) sum[i] += sum[i - 1];
      long long Min = sum[2];
      for (int i = 2; i <= 2 * k; ++i) Min = min(sum[i], Min);
      cout << Min << "\n";
    }
  }
} NSPACE;
int main() {
  ;
  ios_base::sync_with_stdio(false);
  cout.tie(0);
  cin.tie(0);
  NSPACE.solve();
}