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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  cin >> t;
  while (t--) {
    int n, k;
    cin >> n >> k;
    vector<int> v(n + 1);
    vector<int> rs(2 * k + 1);
    vector<int> mini;
    vector<int> maxi;
    for (int i = 1; i <= n; i++) {
      cin >> v[i];
      if (v[i] > k) v[i] = 0;
    }
    int c = 0;
    for (int i = 1; i <= n / 2; i++) {
      if (v[i] != 0 && v[n - i + 1] != 0) {
        rs[v[i] + v[n - i + 1]]++;
        mini.push_back(min(v[i], v[n - i + 1]) + 1);
        maxi.push_back(max(v[i], v[n - i + 1]) + k);
      } else if ((v[i] != 0 && v[n - i + 1] != 0))
        c++;
      else if (v[i] == 0) {
        mini.push_back(v[n - i + 1] + 1);
        maxi.push_back(v[n - i + 1] + k);
      } else if (v[n - i + 1]) {
        mini.push_back(v[i] + 1);
        maxi.push_back(v[i] + k);
      }
    }
    int ans = n;
    sort(mini.begin(), mini.end());
    sort(maxi.begin(), maxi.end());
    for (int i = 2; i <= 2 * k; i++) {
      int x = upper_bound(mini.begin(), mini.end(), i) - mini.begin();
      int y = lower_bound(maxi.begin(), maxi.end(), i) - maxi.begin();
      int oc = x - y;
      int tc = mini.size() - x + y;
      ans = min(ans, oc + 2 * tc + c - rs[i]);
    }
    cout << ans << endl;
  }
}