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;
struct edge {
  int dis;
  int st;
  int ed;
  edge(int dis, int st, int ed) {
    this->dis = dis;
    this->st = st;
    this->ed = ed;
  }
  bool operator<(const edge& rhs) const {
    if (dis == rhs.dis) {
      if (st == rhs.st) return ed < rhs.ed;
      return st < rhs.st;
    }
    return dis < rhs.dis;
  }
};
int dx[9] = {0, 1, 0, -1, 0, 1, 1, -1, -1};
int dy[9] = {-1, 0, 1, 0, 0, 1, -1, 1, -1};
int ans = 0;
long long arr[200001];
bool desc(int a, int b) { return a > b; }
int gcd(int a, int b) {
  if (a < b) swap(a, b);
  while (!b == 0) {
    int r = a % b;
    a = b;
    b = r;
  }
  return a;
}
int main() {
  std::ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  for (int test_case = 0; test_case < t; ++test_case) {
    int n, k;
    cin >> n >> k;
    ans = 0;
    int sum[2 * 200001] = {
        0,
    };
    long long sub = 1;
    set<long long> s;
    for (int i = 0; i < n; ++i) {
      cin >> arr[i];
    }
    for (int i = 0; i < n / 2; ++i) {
      int a = arr[i], b = arr[n - i - 1];
      if (a > b) swap(a, b);
      sum[0] += 2;
      sum[a + b]--;
      sum[a + b + 1]++;
      sum[k + b + 1]++;
      sum[a + 1]--;
    }
    ans = 1000000000;
    for (int i = 1; i <= 2 * k; ++i) {
      sum[i] += sum[i - 1];
      ans = min(ans, sum[i]);
    }
    cout << ans << '\n';
  }
  return 0;
}