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 = 1e9 + 7;
const int MAXW = 1e5 + 1;
bool ispal(string s) {
  int start = 0, end = s.size() - 1;
  while (start < end) {
    if (s[start] == s[end]) {
      end--;
      start++;
    } else {
      return false;
    }
  }
  return true;
}
int ans;
vector<int> temp;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long q = 1;
  cin >> q;
  while (q--) {
    int n, k;
    cin >> n >> k;
    vector<int> v(n);
    vector<int> ones_start(k * 2 + 1);
    vector<int> ones_end(k * 2 + 1);
    vector<int> twos_start(k * 2 + 1);
    vector<int> twos_end(k * 2 + 1);
    for (int i = 0; i < n; i++) {
      cin >> v[i];
    }
    for (int i = 0; i < n / 2; i++) {
      int first = v[i];
      int second = v[n - 1 - i];
      int curr = first + second;
      int least_one = min(first, second) + 1;
      int most_one = max(first, second) + k;
      if (curr != 2) ones_start[least_one]++;
      if (curr != 2 * k) ones_start[curr + 1]++;
      if (curr != 2) ones_end[curr - 1]++;
      if (curr != 2 * k) ones_end[most_one]++;
      if (least_one > 2) {
        twos_start[2]++;
        twos_end[least_one - 1]++;
      }
      if (most_one < k * 2) {
        twos_start[most_one + 1]++;
        twos_end[k * 2]++;
      }
    }
    int ans = INT_MAX, curr = 0;
    for (int i = 2; i <= k * 2; i++) {
      curr += ones_start[i] + 2 * twos_start[i];
      ans = min(ans, curr);
      curr -= (ones_end[i] + 2 * twos_end[i]);
    }
    cout << ans;
    cout << endl;
  }
  return 0;
}