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;
using lli = long long;
using lld = long double;
using ulli = unsigned long long int;
using pll = pair<lli, lli>;
using ttt = pair<lli, pll>;
using vttt = vector<ttt>;
using vll = vector<pll>;
using vl = vector<lli>;
using vi = vector<int>;
using vvi = vector<vector<int>>;
using cd = complex<lld>;
void sanjeev(int kkkk) {
  int t;
  cin >> t;
  while (t > 0) {
    t--;
    int n, k;
    cin >> n >> k;
    vector<int> A;
    for (int i = 0; i < n; i++) {
      int tmp;
      cin >> tmp;
      A.push_back(tmp);
    }
    vector<int> frq(2 * k + 1, 0);
    for (int i = 0; i < n / 2; i++) {
      int tmp = A[i] + A[n - 1 - i];
      frq[tmp]++;
    }
    vector<int> prf(2 * k + 1, 0);
    for (int i = 0; i < n / 2; i++) {
      int a = min(A[i], A[n - i - 1]) + 1;
      int b = max(A[i], A[n - 1 - i]) + k;
      prf[a]++;
      prf[b + 1]--;
    }
    for (int i = 1; i < prf.size(); i++) prf[i] += prf[i - 1];
    long long int ans = INT_MAX;
    for (int i = 0; i < prf.size(); i++) {
      long long int tmp = (n / 2 - prf[i]) * 2 + prf[i] - frq[i];
      ans = min(ans, tmp);
    }
    cout << ans << '\n';
  }
}
int main() {
  if (!0) ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cout << fixed << setprecision(10);
  int t = 1;
  for (int i = 0; i < t; i++) {
    sanjeev(i + 1);
  }
  if (0) system("pause");
}