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>
long long a[200009], prefix[2 * 200009];
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL), cout.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    long long n, k, i;
    cin >> n >> k;
    for (i = 1; i <= n; i++) cin >> a[i];
    memset(prefix, 0, sizeof(prefix));
    map<long long, long long> m;
    for (i = 1; i <= n / 2; i++) {
      long long l = a[i];
      long long r = a[n - i + 1];
      if (l > r) swap(l, r);
      m[l + r]++;
      prefix[l + 1]++;
      prefix[r + k + 1]--;
    }
    for (i = 2; i <= 2 * k; i++) {
      prefix[i] += prefix[i - 1];
    }
    long long ans = LONG_LONG_MAX;
    for (i = 2; i <= 2 * k; i++) {
      long long c0 = m[i];
      long long c1 = prefix[i] - c0;
      long long c2 = (n / 2) - c0 - c1;
      ans = min(ans, c1 + (2 * c2));
    }
    cout << ans << endl;
  }
  return 0;
}