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 main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    int n, k;
    scanf("%d%d", &n, &k);
    vector<int> a(n);
    for (int i = 0; i < n; i++) scanf("%d", &a[i]);
    vector<array<int, 2>> v(n / 2);
    vector<int> cnt(2 * k + 1, 0);
    for (int i = 0; i < n / 2; i++) {
      cnt[a[i] + a[n - i - 1]]++;
      v[i] = {min(a[i], a[n - i - 1]) + 1, max(a[i], a[n - i - 1]) + k};
    }
    sort(v.begin(), v.end());
    int res = n;
    priority_queue<int, vector<int>, greater<int>> q;
    int idx = 0;
    for (int x = 1; x <= 2 * k; x++) {
      while (idx < n / 2 && v[idx][0] == x) q.push(v[idx++][1]);
      int one = q.size() - cnt[x];
      int two = n / 2 - cnt[x] - one;
      res = min(res, one + 2 * two);
      while (!q.empty() && q.top() == x) q.pop();
    }
    printf("%d\n", res);
  }
  return 0;
}