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;
const int oo = 1e9 + 9;
const long long inf = 1e18 + 18;
const int max6 = 1e6 + 6;
const int modx = 1e9 + 123;
const int mody = 997;
const int base = 137;
int a[max6], dd[max6], cnt[max6];
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n, k;
    cin >> n >> k;
    for (int i = 0; i <= 2 * k; ++i) dd[i] = cnt[i] = 0;
    for (int i = 1; i <= n; ++i) cin >> a[i];
    int m = n / 2;
    for (int i = 1; i <= m; ++i) {
      int gmax = max(a[i], a[n - i + 1]) + k;
      int gmin = min(a[i], a[n - i + 1]) + 1;
      dd[gmin]++;
      dd[gmax + 1]--;
      cnt[a[i] + a[n - i + 1]]++;
    }
    int res = oo;
    dd[1] += dd[0];
    for (int i = 2; i <= 2 * k; ++i) {
      dd[i] += dd[i - 1];
      res = min(res, n - dd[i] - cnt[i]);
    }
    cout << res << "\n";
  }
}