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>
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("-O3")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  ;
  int t;
  cin >> t;
  while (t--) {
    int n, k;
    unordered_map<int, int> cnt;
    cin >> n >> k;
    vector<int> a(n);
    vector<int> pref(2 * k + 2, 0);
    for (int i = (0); i < (n); i++) {
      cin >> a[i];
    }
    for (int i = (0); i < (n / 2); i++) {
      int v1 = a[i];
      int v2 = a[n - 1 - i];
      cnt[v1 + v2]++;
      pref[min(v1, v2) + 1] += 1;
      pref[max(v1, v2) + k + 1] -= 1;
    }
    for (int i = (1); i < (pref.size()); i++) {
      pref[i] += pref[i - 1];
    }
    int ans = (int)1e9;
    for (int i = 2; i <= 2 * k; i++) {
      int cur = pref[i] - cnt[i];
      cur += (n / 2 - pref[i]) * 2;
      ans = min(ans, cur);
    }
    cout << ans << "\n";
  }
  return 0;
}