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;
const long long MOD = 1000000007;
const long long N = 1000;
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long test;
  cin >> test;
  while (test--) {
    long long n;
    cin >> n;
    long long k;
    cin >> k;
    long long _N = 3 * k;
    pair<long long, long long> diff[_N];
    memset(diff, 0, sizeof diff);
    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++) {
      long long left = a[i];
      long long right = a[n - i - 1];
      long long l = 2;
      long long r = 2 * k;
      diff[l].first += 2;
      diff[r + 1].first -= 2;
      diff[l].second += 1;
      diff[r + 1].second -= 1;
      l = min(left, right) + 1;
      r = max(left, right) + k;
      diff[l].first--;
      diff[r + 1].first++;
      diff[left + right].first--;
      diff[left + right + 1].first++;
    }
    for (long long i = 1; i < _N; i++) {
      diff[i].first += diff[i - 1].first;
      diff[i].second += diff[i - 1].second;
    }
    long long res = LONG_MAX;
    for (long long i = 0; i < _N; i++) {
      if (diff[i].second == n / 2) {
        res = min(res, diff[i].first);
      }
    }
    cout << res << '\n';
  }
  return 0;
}