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 <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  std::cerr << name << " : " << arg1 << '\n';
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  std::cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
template <typename T, typename U>
static inline void amin(T& x, U y) {
  if (y < x) x = y;
}
template <typename T, typename U>
static inline void amax(T& x, U y) {
  if (x < y) x = y;
}
long long max(long long a, long long b) { return (a > b) ? a : b; }
long long min(long long a, long long b) { return (a < b) ? a : b; }
const long long N = 4e5 + 10;
long long n;
long long t[2 * N];
void build() {
  for (long long i = n - 1; i >= 0; i--) t[i] = (t[i << 1] + t[i << 1 | 1]);
}
void modify(long long l, long long r, long long val) {
  if (l >= r) return;
  for (l += n, r += n; l < r; l >>= 1, r >>= 1) {
    if (l & 1) t[l++] += val;
    if (r & 1) t[--r] += val;
  }
}
long long query(long long p) {
  long long res = 0;
  for (p += n; p > 0; p >>= 1) res += t[p];
  return res;
}
long long solve() {
  long long n;
  cin >> n;
  long long k;
  cin >> k;
  ::n = 2 * k + 1;
  long long a[n];
  for (long long i = 0; i < ::n; i++) t[i + ::n] = 0;
  build();
  for (long long i = 0; i < n; ++i) cin >> a[i];
  for (long long i = 0; i < n / 2; i++) {
    long long f = min(a[i], a[n - i - 1]), s = max(a[i], a[n - i - 1]);
    modify(f + 1, s + k + 1, 1);
    modify(f + s, f + s + 1, -1);
    modify(2, f + 1, 2);
    modify(s + k + 1, 2 * k + 1, 2);
  }
  long long ans = n - 1;
  for (long long i = 2; i <= 2 * k; i++) {
    amin(ans, query(i));
  }
  cout << ans << '\n';
  ;
  return 0;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}