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;
template <class _T>
inline void read(_T &_a) {
  bool f = 0;
  char _c = getchar();
  _a = 0;
  while (_c < '0' || _c > '9') {
    if (_c == '-') f = 1;
    _c = getchar();
  }
  while (_c >= '0' && _c <= '9') {
    _a = (_a << 3) + (_a << 1) - '0' + _c;
    _c = getchar();
  }
  if (f) _a = -_a;
}
int T, n, sum1, sum2;
int main() {
  for (read(T); T; --T) {
    read(n);
    sum1 = sum2 = 0;
    if ((n / 2) & 1) {
      printf("NO\n");
      continue;
    };
    printf("YES\n");
    for (register int i = 2; i <= n; i += 2) printf("%d ", i), sum1 += i;
    n >>= 1;
    for (register int i = 1; i < n; ++i)
      printf("%d ", 2 * i - 1), sum2 += 2 * i - 1;
    printf("%d\n", sum1 - sum2);
  }
  return 0;
}