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;
const int MOD = 1000 * 1000 * 1000 + 7;
const int MAXN = 1000 * 100 + 1;
const int INF = INT_MAX;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int t = 1;
  cin >> t;
  while (t--) {
    int n, k;
    cin >> n >> k;
    vector<int> a(n);
    map<int, int> mp;
    for (int &x : a) cin >> x;
    vector<int> pre(2 * k + 5, 0);
    for (int i = 0; 2 * i < n; ++i) {
      mp[a[i] + a[n - i - 1]]++;
      int mx = a[i], mn = a[n - i - 1];
      if (mx < mn) swap(mx, mn);
      pre[mn + 1] += 1;
      pre[mx + k + 1] += -1;
    }
    for (int i = 1; i < 2 * k + 5; ++i) pre[i] += pre[i - 1];
    int res = INT_MAX;
    for (int i = 2; i <= 2 * k; ++i) {
      int cnt0 = mp[i];
      int cnt1 = pre[i] - cnt0;
      int cnt2 = n / 2 - cnt1 - cnt0;
      res = min(res, cnt1 + 2 * cnt2);
    }
    cout << res, cout << "\n";
  }
  return 0;
}
