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() {
  int t = 1;
  cin >> t;
  while (t--) {
    int n, k;
    cin >> n >> k;
    vector<int> a(n);
    for (auto &i : a) cin >> i;
    vector<int> cnt(2 * k + 1);
    for (int i = 0; i < n / 2; i++) {
      ++cnt[a[i] + a[n - i - 1]];
    }
    vector<int> pref(2 * k + 2);
    for (int i = 0; i < n / 2; i++) {
      int l1 = a[i] + 1, r1 = a[i] + k;
      int l2 = a[n - i - 1] + 1, r2 = a[n - i - 1] + k;
      assert(max(l1, l2) <= min(r1, r2));
      ++pref[min(l1, l2)];
      --pref[max(r1, r2) + 1];
    }
    for (int i = 1; i < 2 * k + 1; i++) pref[i] += pref[i - 1];
    int ans = 1e9;
    for (int sum = 2; sum <= 2 * k; sum++) {
      ans = min(ans, (pref[sum] - cnt[sum]) + (n / 2 - pref[sum]) * 2);
    }
    cout << ans << "\n";
  }
  return 0;
}
