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>
const long long int inf = (1e9 + 3);
const int M97 = ((int)1e9 + 7);
using namespace std;
long long int i, j, k;
void solve() {
  int n;
  cin >> n >> k;
  int a[n];
  for (int zzz = 0; zzz < n; zzz++) {
    cin >> a[zzz];
  }
  vector<int> ones(2 * k + 3), zeroes(2 * k + 3);
  for (i = 0; i < n / 2; i++) {
    pair<int, int> x =
        make_pair(min(a[i], a[n - 1 - i]) + 1, max(a[i], a[n - i - 1]) + k);
    zeroes[a[i] + a[n - i - 1]]++;
    ones[x.first]++;
    ones[x.second + 1]--;
  }
  int one = 0;
  int ans = inf;
  for (i = 2; i < 2 * k + 1; i++) {
    one += ones[i];
    int local = one - zeroes[i];
    local += ((n / 2 - one) << 1);
    ans = min(ans, local);
  }
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int T;
  cin >> T;
  while (T--) solve();
  return 0;
}