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 main() {
  long long int n, t;
  scanf("%lld", &t);
  while (t--) {
    scanf("%lld", &n);
    vector<int> arr;
    long long int cnt = 0;
    long long int d = n / 2;
    if (d % 2 != 0) {
      printf("NO\n");
    } else {
      printf("YES\n");
      long long int j = 0, sum = 0;
      cnt = 0;
      for (int k = 2;; k += 2) {
        if (cnt < d) {
          arr.push_back(k);
          cnt++;
          j++;
          sum += k;
        }
        if (cnt == d) break;
      }
      cnt = 0;
      for (int i = 1; i < sum; i += 2) {
        if (cnt < d) {
          arr.push_back(i);
          sum -= i;
          cnt++;
          if ((cnt == d - 1) && (sum % 2 != 0)) {
            arr.push_back(sum);
            cnt++;
            break;
          } else if ((cnt == d - 1) && (sum % 2 == 0)) {
            arr.push_back(i + 2);
            i = i + 2;
            sum = sum - 2;
          }
        }
        j++;
      }
      for (int l = 0; l < arr.size(); l++) cout << arr[l] << " ";
      printf("\n");
    }
  }
  return 0;
}