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;
void solve() {
  int n, k;
  cin >> n >> k;
  int a[n];
  for (int i = 0; i < n; i++) cin >> a[i];
  int pre[2 * k + 11];
  for (int i = 0; i <= 2 * k + 10; i++) pre[i] = 0;
  unordered_map<int, int> mp;
  vector<pair<int, int>> v;
  for (int i = 0; i < n / 2; i++) {
    v.push_back({min(a[i], a[n - i - 1]), max(a[i], a[n - i - 1])});
    mp[a[i] + a[n - i - 1]]++;
  }
  for (auto it : v) {
    pre[it.first + 1]++;
    pre[it.second + k + 1]--;
  }
  int sum = 0;
  for (int i = 0; i <= 2 * k + 10; i++) {
    sum += pre[i];
    pre[i] = sum;
  }
  int ans = n;
  for (int i = 2; i <= 2 * k; i++) {
    int c1 = mp[i];
    int c2 = pre[i] - c1;
    int c3 = (n / 2) - c1 - c2;
    ans = min(ans, c2 + (2 * c3));
  }
  cout << ans << endl;
  return;
}
int main() {
  int t;
  cin >> t;
  while (t--) solve();
  return 0;
}