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 n, k;
int mi_solve(int x, vector<int>& v) {
  int sz = v.size();
  int lo = 0;
  int hi = sz - 1;
  int ans = -1;
  while (lo <= hi) {
    int mid = lo + (hi - lo) / 2;
    if (v[mid] >= x) {
      ans = mid;
      hi = mid - 1;
    } else
      lo = mid + 1;
  }
  return ans == -1 ? 0 : sz - ans;
}
int ma_solve(int x, vector<int>& v) {
  int lo = 0;
  int hi = v.size() - 1;
  int ans = -1;
  while (lo <= hi) {
    int mid = lo + (hi - lo) / 2;
    if (v[mid] < x) {
      ans = mid;
      lo = mid + 1;
    } else
      hi = mid - 1;
  }
  return ans + 1;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int T;
  cin >> T;
  while (T--) {
    cin >> n >> k;
    vector<int> a(n + 1);
    for (int i = 1; i <= n; i++) cin >> a[i];
    vector<int> m(2 * k + 1, 0);
    vector<int> mi, ma;
    for (int i = 1; i <= n / 2; i++) {
      m[a[i] + a[n - i + 1]] += 1;
      mi.push_back(min(a[i], a[n - i + 1]));
      ma.push_back(max(a[i], a[n - i + 1]));
    }
    sort(mi.begin(), mi.end());
    sort(ma.begin(), ma.end());
    int ans = 2 * n + 1;
    for (int i = 2; i <= k; i++) {
      int curr = n;
      curr -= 2 * m[i];
      curr /= 2;
      int d = mi_solve(i, mi);
      curr += d;
      ans = min(ans, curr);
    }
    for (int i = k + 1; i <= 2 * k; i++) {
      int curr = n;
      curr -= 2 * m[i];
      curr /= 2;
      int d = ma_solve(i - k, ma);
      curr += d;
      ans = min(ans, curr);
    }
    cout << ans << endl;
  }
  return 0;
}