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;
template <class T>
ostream &operator<<(ostream &os, const vector<T> &p) {
  os << "[";
  for (auto &it : p) os << it << " ";
  return os << "]";
}
template <class S, class T>
ostream &operator<<(ostream &os, const pair<S, T> &p) {
  return os << "(" << p.first << "," << p.second << ")";
}
template <class T>
bool umin(T &a, T b) {
  return a > b ? (a = b, true) : false;
}
template <class T>
bool umax(T &a, T b) {
  return a < b ? (a = b, true) : false;
}
const long long N = 2e5 + 10;
void solve() {
  long long n;
  cin >> n;
  long long k;
  cin >> k;
  vector<long long> sum(2 * k + 1), mx(2 * k + 1), mn(2 * k + 1);
  vector<long long> a(n);
  for (long long i = 0; i < n; i++) cin >> a[i];
  for (long long i = 0; i < n / 2; i++) {
    mx[max(a[i], a[n - 1 - i])]++;
    mn[min(a[i], a[n - 1 - i])]++;
    sum[a[i] + a[n - 1 - i]]++;
  }
  for (long long i = 2; i <= 2 * k; i++) mx[i] += mx[i - 1];
  for (long long i = 2 * k; i >= 2; i--) mn[i - 1] += mn[i];
  long long ans = n;
  for (long long i = 2; i <= 2 * k; i++) {
    long long two = mn[i] + ((i - k - 1) >= 1 ? mx[i - k - 1] : 0LL);
    long long zero = sum[i];
    long long one = n / 2 - two - zero;
    umin(ans, 2 * two + one);
  }
  cout << ans << endl;
}
signed main() {
  ios_base::sync_with_stdio(false), cin.tie(nullptr);
  long long q = 1;
  cin >> q;
  while (q--) {
    solve();
  }
  return 0;
}