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) {
  if (!a) return b;
  return gcd(b % a, a);
}
long long mpower(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) {
      res = (res * x) % p;
    }
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long power(long long x, long long y) {
  long long res = 1;
  while (y > 0) {
    if (y & 1) res = res * x;
    y = y >> 1;
    x = x * x;
  }
  return res;
}
long long choose(long long n, long long k) {
  if (k == 0) return 1;
  if (k == 1) return n % 998244353;
  return (((n * choose(n - 1, k - 1)) % 998244353) *
          mpower(k, 998244353 - 2, 998244353)) %
         998244353;
}
void precal() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout << fixed;
  cout << setprecision(9);
  srand(time(NULL));
}
int main() {
  precal();
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    long long t1 = n / 2;
    if (t1 % 2 != 0) {
      cout << "NO" << endl;
    } else {
      cout << "YES" << endl;
      for (long long i = 0; i < t1; i++) {
        cout << 2 * (i + 1) << " ";
      }
      for (long long i = 0; i < t1 - 1; i++) {
        cout << 2 * i + 1 << " ";
      }
      cout << t1 * (t1 + 1) - (t1 - 1) * (t1 - 1) << endl;
    }
  }
}