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 setBit(int num, int idx, int value) {
  return (value) ? (num | (1 << idx)) : (num & ~(1 << idx));
}
int getBit(int num, int idx) { return ((num >> idx) & 1) == 1; }
int cntBit(int num) {
  int ret;
  while (num) {
    ret += num % 2;
    num /= 2;
  }
  return ret;
}
void run() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr), cout.tie(nullptr);
}
int n, need;
void solve() {
  int n, k;
  cin >> n >> k;
  vector<int> cnt(2 * k + 2);
  vector<int> given(2 * k + 3);
  vector<int> a, b;
  for (int i = 0; i < n / 2; ++i) {
    int el;
    cin >> el;
    a.push_back(el);
  }
  for (int i = 0; i < n / 2; ++i) {
    int el;
    cin >> el;
    b.push_back(el);
  }
  reverse(((b).begin()), ((b).end()));
  for (int i = 0; i < n / 2; ++i) {
    cnt[max(a[i] + k, b[i] + k) + 1]--;
    cnt[min(1 + b[i], 1 + a[i])]++;
    given[a[i] + b[i]]++;
  }
  for (int i = 1; i < 2 * k + 2; ++i) {
    cnt[i] += cnt[i - 1];
  }
  int ans = 1e9;
  for (int i = 2; i <= 2 * k + 1; ++i) {
    int ab = i;
    int cur = cnt[ab] + 2 * (n / 2 - cnt[ab]) - given[ab];
    ans = min(cur, ans);
  }
  cout << ans << endl;
}
int main() {
  run();
  int t = 1;
  cin >> t;
  for (int tt = 0; tt < t; ++tt) {
    solve();
  }
  return 0;
}