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 double PI =
    3.14159265358979323846264338327950288419716939937510582097494459230781640628620899863;
const long long INF = 1e9 + 7;
long long pw(long long a, int b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = (res * a) % INF;
    a = (a * a) % INF;
    b >>= 1;
  }
  return res;
}
void edit() { return; }
void solve() {
  int n, k;
  cin >> n >> k;
  int a[n], ans = 100000000;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  vector<int> v1, v2;
  map<int, int> mp;
  map<int, int>::iterator it;
  for (int i = 0; i < n / 2; i++) {
    v1.push_back(min(a[i], a[n - i - 1]));
    v2.push_back(max(a[i], a[n - i - 1]));
    mp[a[i] + a[n - i - 1]]++;
  }
  sort(v1.begin(), v1.end());
  sort(v2.begin(), v2.end());
  for (int i = 2; i < 2 * k + 1; i++) {
    int p1 = lower_bound(v1.begin(), v1.end(), i) - v1.begin();
    int p2 = lower_bound(v2.begin(), v2.end(), i - k) - v2.begin();
    int x = n + p2 - p1 - mp[i];
    ans = min(ans, x);
  }
  cout << ans << "\n";
  return;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}