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;
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int T;
  cin >> T;
  while (T--) {
    int N, K;
    cin >> N >> K;
    int a[N];
    for (int i = 0; i < N; i++) {
      cin >> a[i];
    }
    int b[2 * K + 3], c[2 * K + 3];
    memset(b, 0, sizeof b);
    memset(c, 0, sizeof c);
    for (int i = 0; i < N / 2; i++) {
      b[min(a[i], a[N - (i + 1)]) + 1] += 1;
      b[max(a[i], a[N - (i + 1)]) + K + 1] -= 1;
      c[a[i] + a[N - (i + 1)]]++;
    }
    int d[2 * K + 3];
    memset(d, 0, sizeof d);
    for (int i = 0; i < 2 * K + 3; i++) {
      if (i == 0)
        d[i] = b[i];
      else
        d[i] = d[i - 1] + b[i];
    }
    int ans = INT_MAX;
    for (int i = 2; i <= 2 * K; i++) {
      int t1 = c[i];
      int t2 = d[i];
      ans = min(ans, (t2 - t1) + (+N - (2 * t2)));
    }
    cout << ans << '\n';
  }
  return 0;
}
