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 logb(int base, int x) { return (log(x) / log(base)); }
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int occurs(vector<long long> v, int n) {
  auto it = lower_bound(((v)).begin(), ((v)).end(), (n));
  auto it1 = upper_bound(((v)).begin(), ((v)).end(), (n));
  int x = it1 - it;
  return x;
}
const int mod = 1e9 + 7;
int powm(int a, int b) {
  int res = 1;
  while (b) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return res;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    if ((n / 2) % 2) {
      cout << "NO" << '\n';
      continue;
    }
    cout << "YES" << '\n';
    int j = 2;
    for (int(i) = 0; (i) < (n / 2); ++(i)) {
      cout << j << ' ';
      j += 2;
    }
    j = 1;
    for (int(i) = 0; (i) < (n / 4); ++(i)) {
      cout << j << ' ';
      j += 2;
    }
    j += 2;
    for (int(i) = 0; (i) < (n / 4); ++(i)) {
      cout << j << ' ';
      j += 2;
    }
    cout << '\n';
  }
  return 0;
}