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;
void __print(int x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T &x) {
  int f = 0;
  cerr << '{';
  for (auto &i : x) cerr << (f++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
template <typename... T>
void read(T &...args) {
  ((cin >> args), ...);
}
template <typename... T>
void writ(T &&...args) {
  ((cout << args << " "), ...);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    long long n, k;
    read(n, k);
    vector<long long> inp(n);
    for (long long i = 0; i < n; i++) cin >> inp[i];
    vector<long long> pre((2 * k) + 10, 0);
    vector<pair<long long, long long>> vec;
    map<long long, long long> mp;
    for (long long i = 0; i < n / 2; i++) {
      vec.push_back(make_pair((long long)min(inp[i], inp[n - i - 1]),
                              (long long)max(inp[i], inp[n - i - 1])));
      long long l = vec[i].first + 1;
      long long r = vec[i].second + k;
      mp[inp[i] + inp[n - i - 1]]++;
      pre[l]++;
      pre[r + 1] -= 1;
    }
    long long s = 0;
    for (long long i = 0; i < (2 * k) + 10; i++) {
      s += pre[i];
      pre[i] = s;
    }
    long long mi = n;
    for (long long i = 2; i <= (2 * k); i++) {
      long long ct0 = mp[i];
      long long ct1 = pre[i] - ct0;
      long long ct2 = (n / 2) - ct1 - ct0;
      long long ct = ct1 + 2 * ct2;
      mi = min(mi, ct);
    }
    writ(mi);
    cout << endl;
  }
  return 0;
}