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;
const long long int INF = LLONG_MAX;
vector<long long int> v1;
vector<pair<long long int, long long int> > v2;
string s1, s2, s3;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int a, b, c, d, e, f, i, j, k, cnt1 = 0, cnt2 = 0, sum1 = 0,
                                           sum2 = 0, chk1 = 0, chk2 = 0, T;
  cin >> T;
  while (T--) {
    cnt1 = 0;
    v1.clear();
    cin >> a >> b;
    vector<long long int> ara(b + b + 3, 0);
    for (i = 0; i < a; i++) {
      cin >> c;
      v1.push_back(c);
    }
    ara[0] = a;
    for (i = 0; i < a / 2; i++) {
      long long int mx = max(v1[i], v1[a - i - 1]);
      long long int mn = min(v1[i], v1[a - i - 1]);
      ara[mn + 1]--;
      ara[mx + mn]--;
      ara[mx + mn + 1]++;
      ara[mx + b + 1]++;
    }
    long long int ans = INT_MAX;
    for (i = 1; i <= b + b + 2; i++) {
      ara[i] += ara[i - 1];
      ans = min(ara[i], ans);
    }
    cout << ans << endl;
  }
  return 0;
}