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 = 400100;
long long arr[maxn];
long long cnt[maxn];
long long n, k;
void solve() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> arr[i];
  }
  vector<long long> v, vmin;
  for (int i = 1; i <= n / 2; i++) {
    v.push_back(max(arr[i], arr[n - i + 1]));
    vmin.push_back(min(arr[i], arr[n - i + 1]));
    cnt[arr[i] + arr[n - i + 1]]++;
  }
  sort(v.begin(), v.end());
  sort(vmin.begin(), vmin.end());
  long long result = LLONG_MAX;
  for (int i = 1; i <= 2 * k; i++) {
    long long cnt1 = 0;
    long long ind = vmin.size() - 1;
    for (int cekor = n / 2; cekor > 0; cekor /= 2) {
      while (ind - cekor >= 0 && vmin[ind - cekor] >= i) ind -= cekor;
    }
    if (vmin[ind] >= i) cnt1 = (v.size() - ind);
    long long ind2 = 0;
    for (int cekor = n / 2; cekor > 0; cekor /= 2) {
      while (ind2 + cekor < v.size() && v[ind2 + cekor] + k < i) ind2 += cekor;
    }
    if (v[ind2] + k < i) cnt1 += (ind2 + 1);
    long long curr = n / 2 - cnt[i] + cnt1;
    result = min(result, curr);
  }
  cout << result << "\n";
  for (int i = 1; i <= 2 * k; i++) {
    cnt[i] = 0;
  }
}
int main() {
  int T;
  cin >> T;
  while (T--) {
    solve();
  }
}