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;
void usain_bolt() {
  ios::sync_with_stdio(false);
  cin.tie(0);
}
const int N = 2e5 + 5;
int a[N], sum[2 * N];
int main() {
  usain_bolt();
  int tt;
  cin >> tt;
  for (; tt; --tt) {
    int n, k;
    cin >> n >> k;
    for (int i = 1; i <= n; ++i) cin >> a[i];
    for (int i = 1; i <= n / 2; ++i) {
      int c = a[i], d = a[n - i + 1];
      int mn = min(c, d);
      int mx = max(c, d);
      sum[1] += 2;
      sum[mn + 1] -= 2;
      sum[mn + 1] += 1;
      sum[mn + mx] -= 1;
      sum[mn + mx + 1] += 1;
      sum[mx + k + 1] -= 1;
      sum[mx + k + 1] += 2;
      sum[2 * k + 1] -= 2;
    }
    int cnt = 0, min_sum = 2e9;
    for (int i = 1; i <= 2 * k; ++i) {
      cnt += sum[i];
      min_sum = min(min_sum, cnt);
    }
    cout << min_sum << "\n";
    memset(sum, 0, sizeof(sum));
  }
  return 0;
}
