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;
int ara[400010];
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n, k;
    cin >> n >> k;
    for (int i = 0; i <= 2 * k + 1; i++) ara[i] = 0;
    int now[n + 5];
    for (int i = 1; i <= n; i++) scanf("%d", &now[i]);
    for (int i = 1; i <= n / 2; i++) {
      int x = now[i];
      int y = now[n - i + 1];
      int mx = max(x + k, y + k);
      int mn = min(x + 1, y + 1);
      int p = x + y;
      ara[p + 1]++, ara[mx + 1]--;
      ara[mn]++, ara[p]--;
      ara[1] += 2;
      ara[mn] -= 2;
      ara[mx + 1] += 2;
      ara[2 * k + 1] -= 2;
    }
    int mn2 = 1e9;
    for (int i = 2; i <= 2 * k; i++) {
      ara[i] += ara[i - 1];
      mn2 = min(ara[i], mn2);
    }
    cout << mn2 << endl;
  }
}