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 bs(vector<int> &v, int l, int r, int val) {
  int ans = -1;
  while (l <= r) {
    int m = l + (r - l) / 2;
    if (v[m] < val) {
      ans = m;
      l = m + 1;
    } else
      r = m - 1;
  }
  return ans;
}
int bs2(vector<int> &v, int l, int r, int val) {
  int ans = -1;
  while (l <= r) {
    int m = l + (r - l) / 2;
    if (v[m] > val) {
      ans = m;
      r = m - 1;
    } else
      l = m + 1;
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0);
  int t;
  cin >> t;
  int n, k;
  while (t--) {
    cin >> n >> k;
    int a[n];
    for (int i = 0; i < n; i++) cin >> a[i];
    int fre[400010] = {0};
    for (int i = 0; i < n / 2; i++) fre[a[i] + a[n - 1 - i]]++;
    vector<vector<int> > sum(2, vector<int>(n / 2));
    for (int i = 0; i < n / 2; i++) sum[0][i] = max(a[i], a[n - 1 - i]) + k;
    for (int i = 0; i < n / 2; i++) sum[1][i] = min(a[i], a[n - 1 - i]) + 1;
    sort(sum[0].begin(), sum[0].end());
    sort(sum[1].begin(), sum[1].end());
    int ans = 1000000;
    for (int i = 2; i <= 2 * k; i++) {
      int pos1 = bs(sum[0], 0, n / 2 - 1, i);
      int pos2 = bs2(sum[1], 0, n / 2 - 1, i);
      if (pos1 == -1)
        pos1 = 0;
      else
        pos1++;
      if (pos2 == -1)
        pos2 = 0;
      else
        pos2 = n / 2 - pos2;
      ans = min(ans, pos1 * 2 + pos2 * 2 + (n / 2 - pos1 - pos2 - fre[i]));
    }
    cout << ans << '\n';
  }
  return 0;
}
