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;
const int MXN = (int)4e5 + 7;
const int MOD = (int)998244353;
int n, a[MXN], mod[MXN], k;
void add_mn(int a, int b) {
  int sum = a + b;
  int l = sum - (b - 1);
  mod[l]++;
  mod[sum]--;
  if (l > 2) {
    mod[2] += 2;
    mod[l] -= 2;
  }
}
void add_mx(int a, int b) {
  int sum = a + b;
  int r = (k - a) + sum;
  mod[sum + 1]++;
  mod[r + 1]--;
  mod[r + 1] += 2;
}
void solve() {
  cin >> n >> k;
  fill(mod + 2, mod + (2 * k) + 1, 0);
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (int i = 1; i <= n / 2; i++) {
    if (a[i] > a[n - i + 1]) swap(a[i], a[n - i + 1]);
  }
  for (int i = 1; i <= n / 2; i++) {
    add_mx(a[i], a[n - i + 1]);
    add_mn(a[i], a[n - i + 1]);
  }
  int sum = 0, mn = MOD;
  for (int i = 2; i <= 2 * k; i++) {
    sum += mod[i];
    mn = min(mn, sum);
  }
  cout << mn << endl;
}
signed main() {
  int t;
  ios_base::sync_with_stdio(0);
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}