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;
template <class T1, class T2, class T3 = hash<T1>>
using umap = unordered_map<T1, T2, T3>;
template <class T>
using uset = unordered_set<T>;
template <class T>
using vec = vector<T>;
const long long infll = numeric_limits<long long>::max() >> 1;
const long long inf = numeric_limits<long long>::max() >> 1;
const long long N = 4e5 + 5;
long long n, k, ans, tc;
long long arr[N];
long long cnt[N];
long long mxm[N];
long long mnm[N];
void clear() {
  for (long long i = 1; i <= 2 * k; ++i) {
    cnt[i] = 0;
    mxm[i] = 0;
    mnm[i] = 0;
  }
}
void input() {
  cin >> n >> k;
  for (long long i = 1; i <= n; ++i) {
    cin >> arr[i];
  }
  for (long long i = 1; i <= n / 2; ++i) {
    cnt[arr[i] + arr[n - i + 1]]++;
  }
  for (long long i = 1; i <= n / 2; ++i) {
    long long mx = max(arr[i], arr[n - i + 1]) + k;
    long long mn = min(arr[i], arr[n - i + 1]) + 1;
    mnm[mn]++;
    mxm[mx]++;
  }
  for (long long i = 2 * k; i >= 0; --i) {
    mnm[i] += mnm[i + 1];
  }
  for (long long i = 1; i <= 2 * k; ++i) {
    mxm[i] += mxm[i - 1];
  }
}
void solve() {
  ans = inf;
  for (long long i = 2; i <= 2 * k; ++i) {
    long long c = n / 2 - cnt[i];
    long long x = mnm[i + 1];
    long long y = mxm[i - 1];
    ans = min(ans, c + x + y);
  }
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  cin >> tc;
  while (tc--) {
    clear();
    input();
    solve();
  }
}