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 namespace std;
const long long int M = 1e9 + 7;
void solve() {
  long long int n, k;
  cin >> n >> k;
  vector<long long int> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  long long int mine = M, maxe = -M;
  map<long long int, long long int> m;
  vector<long long int> pre(2 * k + 5);
  for (int i = 0; i < n / 2; i++) {
    maxe = max(maxe, min(a[i], a[n - 1 - i]) + 1);
    mine = min(mine, max(a[i], a[n - 1 - i]) + k);
    long long int x = a[i] + a[n - 1 - i];
    pre[min(a[i], a[n - 1 - i]) + 1]++;
    pre[max(a[i], a[n - 1 - i]) + k + 1]--;
    m[x]++;
  }
  for (int i = 1; i <= 2 * k + 1; i++) {
    pre[i] += pre[i - 1];
  }
  long long int ans = M;
  for (int i = 2; i <= 2 * k; i++) {
    ans = min(ans, (pre[i] - m[i]) + (n / 2 - pre[i]) * 2);
  }
  cout << ans << endl;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  srand(time(NULL));
  long long int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}