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() {
  std::ios::sync_with_stdio(false);
  int T;
  cin >> T;
  while (T--) {
    int n, k;
    cin >> n >> k;
    int a[n];
    for (int i = 0; i < n; i++) {
      cin >> a[i];
    }
    vector<int> pre(4 * k + 1, 0);
    map<long long int, long long int> mp;
    for (int i = 0; i < n / 2; i++) {
      long long int r = max(a[i], a[n - 1 - i]) + k;
      long long int l = min(a[i], a[n - 1 - i]) + 1;
      pre[l]++;
      pre[r + 1] -= 1;
      long long int sum = a[i] + a[n - 1 - i];
      mp[sum]++;
    }
    long long int ans = 1000000000000000001;
    ;
    for (long long int i = 2; i <= 2 * k; i++) {
      pre[i] = pre[i] + pre[i - 1];
    }
    for (int i = 2; i <= 2 * k; i++) {
      long long int j;
      j = pre[i] - mp[i] + 2 * ((n / 2) - pre[i]);
      ans = min(ans, j);
    }
    cout << ans << "\n";
  }
  return 0;
}