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 gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
const long long mod = 1e9 + 7;
const long long N = 2e5 + 9;
void raushan_kumar() {
  long long d = 0, e = 0, f = 0, g = 0;
  double f1 = 0.0, f2 = 0.0, f3 = 0.0;
  long long flag = 0, ans = 0, res = 0, cnt = 0, sum = 0;
  long long n;
  cin >> n;
  long long m = n / 2;
  long long a[n / 2], b[n / 2];
  if (n % 4 == 0) {
    cout << "YES" << endl;
    long long m = n / 2;
    for (long long i = 0, iend = m; i < iend; i++) {
      a[i] = 2 * (i + 1);
      sum += a[i];
    }
    for (long long i = 0; i < m - 1; i++) {
      b[i] = 2 * (i + 1) - 1;
      cnt += b[i];
    }
    b[m - 1] = sum - cnt;
    for (long long i = 0, iend = m; i < iend; i++) {
      cout << a[i] << " ";
    }
    for (long long i = 0, iend = m; i < iend; i++) {
      cout << b[i] << " ";
    }
    cout << endl;
  } else {
    cout << "NO" << endl;
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  cin >> t;
  while (t--) {
    raushan_kumar();
  }
}