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
#include <bits/stdc++.h>
using namespace std;
const long long int N = 2 * 1e5 + 2;
long long int n, k;
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long int tt;
  cin >> tt;
  for (long long int qq = 1; qq <= tt; ++qq) {
    cin >> n >> k;
    vector<long long int> a(n + 1), cnt(2 * k + 1, 0), temp(2 * k + 1, 0);
    for (long long int i = 1; i <= n; ++i) cin >> a[i];
    for (long long int i = 1; i <= n / 2; ++i) {
      long long int x = a[i], y = a[n + 1 - i];
      if (x > y) swap(x, y);
      cnt[x + y]++;
      long long int mn = x + 1;
      long long int mx = y + k;
      long long int mid = x + y;
      assert(mn <= mid && mid <= mx);
      temp[mn]++;
      temp[mid]--;
      temp[mid + 1]++;
      temp[mx + 1]--;
    }
    for (long long int i = 1; i <= 2 * k + 1; ++i) {
      temp[i] += temp[i - 1];
    }
    long long int res = 1e12;
    for (long long int i = 2; i <= 2 * k; ++i) {
      long long int twos = (n / 2) - cnt[i] - temp[i];
      res = min(res, 2 * twos + temp[i]);
    }
    cout << res << '\n';
  }
  return 0;
}
