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 te, n, k, x, sl, a[200005], kq, dd[400005], vt;
vector<int> vec, vec2;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> te;
  while (te--) {
    cin >> n >> k;
    vec.clear();
    vec2.clear();
    for (int i = 1; i <= n; ++i) cin >> a[i];
    for (int i = 1; i <= n / 2; ++i) {
      dd[a[i] + a[n - i + 1]]++;
      x = k + max(a[i], a[n - i + 1]);
      vec.push_back(x);
      vec2.push_back(min(a[i], a[n - i + 1]));
    }
    sort(vec.begin(), vec.end());
    sort(vec2.begin(), vec2.end());
    kq = n;
    vt = 0;
    for (int i = 1; i <= k; ++i) {
      while (vt < vec2.size() && vec2[vt] < i) vt++;
      sl = (n / 2 - vt) * 2 + vt - dd[i];
      kq = min(kq, sl);
    }
    vt = 0;
    for (int i = k + 1; i <= k * 2; ++i) {
      while (vt < vec.size() && vec[vt] < i) vt++;
      sl = vt * 2 + n / 2 - vt - dd[i];
      kq = min(kq, sl);
    }
    cout << kq << '\n';
    for (int i = 1; i <= n / 2; ++i) dd[a[i] + a[n - i + 1]] = 0;
  }
  return 0;
}