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() {
  int t;
  cin >> t;
  while (t--) {
    int n, k;
    cin >> n >> k;
    int arr[n];
    for (int i = 0; i < n; i++) {
      cin >> arr[i];
    }
    long long int sum[n / 2];
    int j = n / 2;
    int i = j - 1;
    int mini[n / 2];
    int maxii[n / 2];
    for (; i >= 0 && j < n; i--, j++) {
      sum[i] = arr[i] + arr[j];
      mini[i] = min(arr[i], arr[j]);
      maxii[i] = max(arr[i], arr[j]);
    }
    sort(mini, mini + n / 2);
    sort(maxii, maxii + n / 2);
    int ans = INT_MAX;
    map<long long int, int> mp;
    for (int i = 0; i < n / 2; i++) {
      mp[sum[i]]++;
    }
    for (int i = 2; i <= 2 * k; i++) {
      int temp1 = n / 2, temp2 = 0;
      if (i <= k) temp1 = lower_bound(mini, mini + n / 2, i) - mini;
      if (i > k) {
        temp2 = lower_bound(maxii, maxii + n / 2, i - k) - maxii;
      }
      int n1 = n / 2 - temp1;
      int n2 = temp2;
      ans = min(ans, n1 + n2 + n / 2 - mp[i]);
    }
    cout << ans << endl;
  }
  return 0;
}