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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    int n, k;
    cin >> n >> k;
    int a[n];
    for (int i = 0; i < n; i++) cin >> a[i];
    vector<pair<int, int>> v;
    for (int i = 0; i < (n / 2); i++) {
      v.push_back(make_pair(max(a[i], a[n - i - 1]), min(a[n - 1 - i], a[i])));
    }
    int dp[2 * k + 2];
    map<int, int> ans;
    for (int i = 0; i < n / 2; i++) {
      ans[v[i].first + v[i].second] += 1;
    }
    for (int i = 1; i <= 2 * k + 1; i++) dp[i] = 0;
    for (int i = 0; i < n / 2; i++) {
      int m1 = v[i].first + k;
      int m2 = v[i].second + 1;
      dp[m2] += 1;
      dp[m1 + 1] -= 1;
    }
    int ansm = n / 2;
    int s = 0;
    for (int i = 2; i <= 2 * k; i++) {
      s += dp[i];
      ansm = min(ansm, s - ans[i] + 2 * (n / 2 - s));
    }
    cout << ansm << endl;
  }
  return 0;
}