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 const N = 2e5 + 5;
long long a[N];
set<long long> st;
void solve() {
  st.clear();
  int n;
  cin >> n;
  int x = n / 2;
  if (x & 1) {
    cout << "NO" << endl;
    return;
  }
  long long frst = 1;
  st.insert(frst);
  long long ev = 2;
  for (int i = 0; i < x; i += 2) {
    long long sum = ev;
    a[i] = ev;
    ev += 2;
    a[i + 1] = ev;
    sum += ev;
    ev += 2;
    a[x + i] = frst;
    a[x + i + 1] = sum - frst;
    st.insert(frst);
    st.insert(sum - frst);
    while (1) {
      frst += 2;
      if (!st.count(frst)) break;
    }
  }
  cout << "YES" << endl;
  for (int i = 0; i < n; i++) cout << a[i] << " ";
  cout << endl;
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int t;
  cin >> t;
  while (t--) solve();
  return 0;
}