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;
long long int power(long long int x, long long int y,
                    long long int p = 1e9 + 7) {
  long long int res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) {
      res = (res * x) % p;
    }
    x = (x * x) % p;
    y = y >> 1;
  }
  return res;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long int t;
  cin >> t;
  while (t--) {
    long long int i, j, k, n;
    vector<long long int> v;
    cin >> n >> k;
    for (i = 0; i < n; i++) {
      cin >> j;
      v.push_back(j);
    }
    vector<long long int> ans(2 * k + 2);
    map<long long int, long long int> m;
    for (i = 0; i < n / 2; i++) {
      long long int x = min(v[i], v[n - i - 1]);
      long long int y = max(v[i], v[n - i - 1]);
      m[x + y]++;
      ans[x + 1]++;
      ans[y + k + 1]--;
    }
    for (i = 3; i <= 2 * k; i++) {
      ans[i] = ans[i] + ans[i - 1];
    }
    long long int mn = LLONG_MAX;
    long long int sum = 0;
    for (i = 2; i < 2 * k + 1; i++) {
      sum = 0;
      j = ans[i];
      sum += (j - m[i]);
      j = (n / 2) - ans[i];
      sum += (2 * j);
      mn = min(mn, sum);
    }
    cout << mn << endl;
  }
  return 0;
}