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 sum[400006] = {0};
long long Min[400006] = {0};
long long Max[400006] = {0};
void solve() {
  long long n;
  cin >> n;
  long long k;
  cin >> k;
  vector<long long> v(n);
  for (long long i = 0; i < n; i++) cin >> v[i];
  for (long long i = 0; i <= 2 * k + 3; i++) Min[i] = Max[i] = sum[i] = 0;
  for (long long i = 0; i < n / 2; i++) {
    Max[max(v[i], v[n - i - 1])]++;
    Min[min(v[i], v[n - i - 1])]++;
    sum[v[i] + v[n - i - 1]]++;
  }
  for (long long i = 1; i <= k; i++) {
    Min[i] += Min[i - 1];
    Max[i] += Max[i - 1];
  }
  long long ans = 2000000000000000005;
  long long c;
  for (long long i = 2; i <= k + 1; i++) {
    c = Min[i - 1];
    if (c <= n / 2) ans = min(ans, c + 2 * ((n / 2) - c) - sum[i]);
  }
  for (long long i = k + 1; i <= 2 * k; i++) {
    if (i - k - 1 >= 0) c = Max[i - k - 1];
    if (c <= n / 2) ans = min(ans, c * 2 + (n / 2 - c) - sum[i]);
  }
  cout << ans << "\n";
}
signed main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}