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 Nmax = 1000005;
const int INF = 1000000000;
const int MOD = 998244353;
void fix(int &x) {
  if (x >= MOD) {
    x -= MOD;
  }
}
int a[Nmax];
pair<int, int> sums[Nmax];
int main() {
  int t, n, k;
  cin.sync_with_stdio(false);
  cin >> t;
  while (t--) {
    cin >> n >> k;
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
    }
    for (int i = 1; i <= n / 2; i++) {
      sums[i].first = a[i] + a[n - i + 1];
      sums[i].second = a[i];
    }
    n >>= 1;
    sort(sums + 1, sums + n + 1);
    vector<int> marked(2 * k + 1, 0);
    vector<int> ans(2 * k + 1, 0);
    int cnt = 0, nr_smaller = 0, p = 1;
    for (int i = 1; i <= 2 * k; i++) {
      while (p <= n && sums[p].first < i) {
        nr_smaller++;
        int x = sums[p].first - sums[p].second;
        marked[sums[p].first + max(k - sums[p].second, k - x)]++;
        p++;
      }
      ans[i] += cnt + nr_smaller;
      cnt += marked[i];
    }
    cnt = 0;
    p = n;
    int nr_greater = 0, sol = INF;
    vector<int> marked2(2 * k + 1, 0);
    for (int i = 2 * k; i; i--) {
      while (p > 0 && sums[p].first > i) {
        nr_greater++;
        int x = sums[p].first - sums[p].second;
        marked2[sums[p].first - max(sums[p].second - 1, x - 1)]++;
        p--;
      }
      ans[i] += cnt + nr_greater;
      sol = min(sol, ans[i]);
      cnt += marked2[i];
    }
    cout << sol << "\n";
  }
  return 0;
}
