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;
using ll = long long;
using vll = vector<ll>;
using pll = pair<ll, ll>;
ll INF = 1000000000000;
int main() {
  cin.tie(nullptr);
  ios::sync_with_stdio(false);
  ll t;
  cin >> t;
  for (int test = 0; test < t; test++) {
    ll n;
    cin >> n;
    ll k;
    cin >> k;
    n /= 2;
    vector<pll> a(n);
    for (int i = 0; i < n; i++) {
      cin >> a.at(i).first;
    }
    for (int i = 0; i < n; i++) {
      cin >> a.at(n - 1 - i).second;
    }
    vll sum(2 * k + 10, 0);
    for (int i = 0; i < n; i++) {
      ll f = a.at(i).first;
      ll s = a.at(i).second;
      sum.at(f + s) -= 1;
      sum.at(f + s + 1) += 1;
      ll mn = min(f, s) + 1;
      ll mx = max(f, s) + k;
      sum.at(2) += 2;
      sum.at(mn) -= 2;
      sum.at(mn) += 1;
      sum.at(mx + 1) -= 1;
      sum.at(mx + 1) += 2;
    }
    ll ans = INF;
    for (int i = 2; i <= 2 * k; i++) {
      sum.at(i) += sum.at(i - 1);
      ans = min(ans, sum.at(i));
    }
    cout << ans << endl;
  }
}