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 long long mod = 1000000007;
const long long N = 200005;
long long n, arr[N], t, k, poss[2 * N], a, b, ans;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> t;
  while (t--) {
    cin >> n >> k;
    memset(poss, 0, sizeof poss);
    for (long long i = 0; i < n; i++) cin >> arr[i];
    for (long long i = 0, j = n - 1; i < j; i++, j--) {
      a = arr[i], b = arr[j];
      if (2 <= min(a, b)) {
        poss[2] += 2, poss[min(a, b) + 1] -= 2;
      }
      if (min(a, b) + 1 <= max(a, b) + k) {
        poss[min(a, b) + 1]++, poss[max(a, b) + k + 1]--;
      }
      if (max(a, b) + k + 1 <= 2 * k) {
        poss[max(a, b) + k + 1] += 2, poss[2 * k + 1] -= 2;
      }
      poss[a + b]--, poss[a + b + 1]++;
    }
    ans = INT_MAX;
    for (long long i = 2; i <= 2 * k; i++) {
      poss[i] += poss[i - 1];
      ans = min(ans, poss[i]);
    }
    cout << ans << '\n';
  }
}