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 main() {
  int t;
  cin >> t;
  while (t--) {
    int n, k;
    cin >> n >> k;
    int ar[n + 1];
    for (int i = 1; i <= n; i++) cin >> ar[i];
    int sm[k + 1];
    int lr[k + 1];
    for (int i = 0; i <= k; i++) {
      sm[i] = 0;
      lr[i] = 0;
    }
    int sum[2 * k + 1];
    for (int i = 0; i <= 2 * k; i++) sum[i] = 0;
    for (int i = 1; i <= n / 2; i++) {
      sm[min(ar[i], ar[n + 1 - i])]++;
      lr[max(ar[i], ar[n + 1 - i])]++;
      sum[ar[i] + ar[n + 1 - i]]++;
    }
    for (int i = k - 1; i >= 1; i--) sm[i] += sm[i + 1];
    for (int i = 2; i <= k; i++) lr[i] += lr[i - 1];
    int ans = INT_MAX;
    for (int i = 1; i <= 2 * k; i++) {
      int x = i;
      if (x <= k) {
        int ag = (n / 2) - sum[x];
        int y = sm[x];
        int rr = ag - y;
        if (2 * y + rr <= ans) {
          ans = 2 * y + rr;
        }
      } else {
        int ag = (n / 2) - sum[x];
        int y = -1;
        if (x - k > 1)
          y = lr[x - k - 1];
        else
          y = 0;
        int rr = ag - y;
        if (2 * y + rr <= ans) {
          ans = 2 * y + rr;
        }
      }
    }
    cout << ans << endl;
  }
  return 0;
}