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;
long long ans = 1ll;
long long MOD = 1000000007;
std::vector<long long> g[51];
bool visited[51];
void dfs(long long i) {
  visited[i] = true;
  for (long long child : g[i]) {
    if (!visited[child]) {
      ans *= 2;
      dfs(child);
    }
  }
}
long long fact(long long k) {
  if (k == 1) return 1;
  return k * fact(k - 1);
}
bool isSubSequence(string tobechecked, string str2, int m, int n) {
  if (m == 0) return true;
  if (n == 0) return false;
  if (tobechecked[m - 1] == str2[n - 1])
    return isSubSequence(tobechecked, str2, m - 1, n - 1);
  return isSubSequence(tobechecked, str2, m, n - 1);
}
int main() {
  long long t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    int k = n / 2;
    if (k % 2 == 1)
      cout << "NO" << endl;
    else {
      std::vector<int> odd;
      std::vector<int> even;
      int z = k / 2;
      for (int i = 1; i <= z; i++) {
        even.push_back((i - 1) * 6 + 2);
        even.push_back((i - 1) * 6 + 4);
        odd.push_back((i - 1) * 6 + 1);
        odd.push_back((i - 1) * 6 + 5);
      }
      cout << "YES" << endl;
      for (int x : even) cout << x << " ";
      for (int x : odd) cout << x << " ";
      cout << endl;
    }
  }
  return 0;
}