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;
const long long Inf = 1e18, Ninf = -1e18, Mod = 1e9 + 7;
const long double Eps = 1e-6;
const string Alph_lower("abcdefghijklmnopqrstuvwxyz"),
    Alph_upper("ABCDEFGHIJKLMNOPQRSTUVWXYZ"), Digits("0123456789");
void Max_self(long long &original_value, long long new_value) {
  original_value = max(original_value, new_value);
}
void Min_self(long long &original_value, long long new_value) {
  original_value = min(original_value, new_value);
}
void To_lowercase(string &str) {
  for (long long i = 0; i < str.length(); i++) str[i] = tolower((str[i]));
}
void To_uppercase(string &str) {
  for (long long i = 0; i < str.length(); i++) str[i] = toupper(str[i]);
}
bool Is_prime(long long num) {
  if (num == 2) return true;
  for (long long div = 2; div <= sqrt(num); div++)
    if (!(num % div)) return false;
  return true;
}
long long GCD(long long num_1, long long num_2) {
  return num_2 ? GCD(num_2, num_1 % num_2) : num_1;
}
const long long M1 = 2e5 + 5, M2 = 4e5 + 10;
long long t;
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> t;
  while (t--) {
    long long n, k;
    vector<pair<long long, long long>> pair_values;
    long long pair_freq[M2] = {0};
    long long answer = Inf;
    cin >> n >> k;
    long long a[n + 5];
    memset(a, 0, sizeof(a));
    long long diff_array[2 * k + 5];
    memset(diff_array, 0, sizeof(diff_array));
    for (long long i = 1; i <= n; i++) cin >> a[i];
    for (long long i = 1; i <= n / 2; i++) {
      long long pair_value = a[i] + a[n - i + 1];
      if (pair_freq[pair_value] == 0)
        pair_values.push_back(make_pair(0, pair_value));
      pair_freq[pair_value]++;
      diff_array[0]++;
      diff_array[pair_value - max(a[i], a[n - i + 1]) + 1]--;
      diff_array[pair_value + (k - min(a[i], a[n - i + 1])) + 1]++;
    }
    for (long long i = 1; i <= 2 * k; i++) diff_array[i] += diff_array[i - 1];
    for (long long i = 0; i < pair_values.size(); i++) {
      pair_values[i].first = pair_freq[pair_values[i].second];
    }
    for (long long i = 0; i < pair_values.size(); i++) {
      Min_self(answer, ((n / 2) - pair_values[i].first) +
                           diff_array[pair_values[i].second]);
    }
    cout << min(n / 2, answer) << "\n";
  }
  return 0;
}