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;
struct Fenwick {
  vector<long long int> BIT;
  long long int N;
  Fenwick(long long int n) {
    N = 2 * n + 2;
    BIT.resize(N + 10);
  }
  void reset() {
    for (long long int i = 0; i <= N; i++) BIT[i] = 0;
  }
  void add(long long int r, long long int value) {
    for (long long int i = r; i <= N; i += (i & (-i))) BIT[i] += value;
  }
  long long int qry(long long int r) {
    long long int ans = 0;
    for (long long int i = r; i >= 1; i -= (i & (-i))) ans += BIT[i];
    return ans;
  }
};
void solve() {
  long long int n, k;
  cin >> n >> k;
  vector<long long int> v(n);
  for (auto &it : v) {
    cin >> it;
  }
  unordered_map<long long int, long long int> hash;
  Fenwick sg1(2 * k + 1), sg2(2 * k + 1);
  for (long long int i = 0; i < n / 2; i++) {
    long long int sum = v[i] + v[n - i - 1];
    hash[sum]++;
    sg1.add(max(v[i], v[n - i - 1]), 1);
    sg2.add(min(v[i], v[n - i - 1]), 1);
  }
  long long int ans = n;
  for (long long int i = 2; i <= 2 * k; i++) {
    long long int cnt = 0;
    long long int extra = 2 * sg1.qry(i - k - 1ll);
    long long int extra_s = n / 2 - sg2.qry(i - 1);
    extra_s *= 2;
    long long int left = n / 2 - extra / 2 - extra_s / 2;
    cnt = cnt + left + extra_s + extra;
    cnt -= hash[i];
    ans = min(ans, cnt);
  }
  cout << ans << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t;
  cin >> t;
  while (t--) {
    solve();
  }
}