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 MAXN = 2e5 + 5;
int a[MAXN], sum[MAXN << 1];
int cnt[MAXN];
int t, n, k;
int mini[MAXN], maxi[MAXN];
int main() {
  cin >> t;
  for (int i = 1; i <= 100000; i++) {
    i = i;
  }
  while (t--) {
    int ans = 2147483647;
    cin >> n >> k;
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
    }
    memset(mini, 0, sizeof mini);
    memset(maxi, 0, sizeof maxi);
    memset(cnt, 0, sizeof cnt);
    for (int i = 1; i <= (n >> 1); i++) {
      cnt[a[i] + a[n - i + 1]]++;
      mini[i] = min(a[i], a[n - i + 1]);
      maxi[i] = max(a[i], a[n - i + 1]) + k;
    }
    sort(mini + 1, mini + 1 + (n >> 1), greater<int>());
    sort(maxi + 1, maxi + 1 + (n >> 1));
    for (int i = 0; i <= (k << 1); i++) {
      int cur = (n >> 1) - cnt[i];
      int left = 1, right = (n >> 1);
      int res = 0;
      while (left <= right) {
        int mid = (left + right) >> 1;
        if (mini[mid] >= i) {
          res = mid;
          left = mid + 1;
        } else {
          right = mid - 1;
        }
      }
      cur += res;
      res = 0;
      left = 1, right = (n >> 1);
      while (left <= right) {
        int mid = (left + right) >> 1;
        if (maxi[mid] < i) {
          res = mid;
          left = mid + 1;
        } else {
          right = mid - 1;
        }
      }
      cur += res;
      ans = min(ans, cur);
    }
    cout << ans << endl;
  }
  return 0;
}