Problem: You are given an array a consisting of n integers (it is guaranteed that n is even, i.e. divisible by 2). All a_i does not exceed some integer k.

Your task is to replace the minimum number of elements (replacement is the following operation: choose some index i from 1 to n and replace a_i with some integer in range [1; k]) to satisfy the following conditions:

  * after all replacements, all a_i are positive integers not greater than k; 
  * for all i from 1 to n/2 the following equation is true: a_i + a_{n - i + 1} = x, where x should be the same for all n/2 pairs of elements. 



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 first line of the test case contains two integers n and k (2 ≤ n ≤ 2 ⋅ 10^5, 1 ≤ k ≤ 2 ⋅ 10^5) — the length of a and the maximum possible value of some a_i correspondingly. It is guratanteed that n is even (i.e. divisible by 2). The second line of the test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ k), where a_i is the i-th element of a.

It is guaranteed that the sum of n (as well as the sum of k) over all test cases does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5, ∑ k ≤ 2 ⋅ 10^5).

Output

For each test case, print the answer — the minimum number of elements you have to replace in a to satisfy the conditions from the problem statement.

Example

Input


4
4 2
1 2 1 2
4 3
1 2 2 1
8 7
6 1 1 7 6 3 4 6
6 6
5 2 6 1 3 4


Output


0
1
4
2

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
using LL = long long int;
template <class TH>
void _dbg(const char *sdbg, TH h) {
  cerr << sdbg << "=" << h << "\n";
}
template <class TH, class... TA>
void _dbg(const char *sdbg, TH h, TA... a) {
  while (*sdbg != ',') cerr << *sdbg++;
  cerr << "=" << h << ",";
  _dbg(sdbg + 1, a...);
}
template <class T>
ostream &operator<<(ostream &os, vector<T> V) {
  os << "[";
  for (auto vv : V) os << vv << ",";
  return os << "]";
}
template <class L, class R>
ostream &operator<<(ostream &os, pair<L, R> P) {
  return os << "(" << P.st << "," << P.nd << ")";
}
const int MAXN = 1e6 + 6;
LL t[MAXN];
LL n;
LL k;
LL events[MAXN];
void solve() {
  cin >> n >> k;
  map<LL, LL> sum;
  for (LL i = 1; i <= n; ++i) {
    cin >> t[i];
  }
  for (LL i = 1; i <= n / 2; ++i) {
    sum[t[i] + t[n - i + 1]]++;
    LL minn = min(t[i], t[n - i + 1]) + 1;
    LL maks = max(t[i], t[n - i + 1]) + k;
    events[minn]++;
    events[maks + 1]--;
  }
  LL res = n;
  LL actRanges = 0;
  for (LL i = 2; i <= 2 * k + 1; ++i) {
    actRanges += events[i];
    LL newRes = n;
    newRes -= 2 * sum[i];
    newRes = newRes - (actRanges - sum[i]);
    (newRes, i);
    res = min(res, newRes);
    events[i] = 0;
  }
  cout << res << endl;
}
int main() {
  int z;
  cin >> z;
  while (z--) {
    solve();
  }
}