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;
long long n, t, v[200001], k, nr, maxi, mini, val, sum[400002], fv[400002];
int main() {
  int i;
  cin >> t;
  while (t--) {
    maxi = 0;
    nr = 0;
    int ok = 1;
    cin >> n >> k;
    for (i = 2; i <= 2 * k + 1; i++) {
      sum[i] = 0;
      fv[i] = 0;
    }
    mini = 2 * k + 1;
    for (i = 1; i <= n; i++) cin >> v[i];
    for (i = 1; i <= n / 2; i++) {
      fv[v[i] + v[n - i + 1]]++;
    }
    for (i = 1; i <= n / 2; i++) {
      int l1 = v[i] + 1, r1 = v[i] + k;
      int l2 = v[n - i + 1] + 1, r2 = v[n - i + 1] + k;
      sum[min(l1, l2)]++;
      if (max(r1, r2) != 2 * k) sum[max(r1, r2) + 1]--;
    }
    for (i = 1; i <= 2 * k + 1; i++) sum[i] += sum[i - 1];
    long long ans = 1e9;
    for (i = 2; i <= 2 * k; i++)
      ans = min(ans, (sum[i] - fv[i]) + (n / 2 - sum[i]) * 2);
    cout << ans << '\n';
  }
  return 0;
}