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 t, n, k;
vector<int> a;
int x;
map<int, int> l, r, s;
int ans;
int d;
int main() {
  cin >> t;
  while (t--) {
    cin >> n >> k;
    a.assign(n, 0);
    l.clear();
    r.clear();
    s.clear();
    for (int i = 0; i < n; ++i) cin >> a[i];
    for (int i = 0; i < n / 2; ++i) {
      if (!(l.count(min(a[i], a[n - 1 - i]) + 1)))
        l[min(a[i], a[n - 1 - i]) + 1] = 1;
      else
        ++l[min(a[i], a[n - 1 - i]) + 1];
      if (!(r.count(max(a[i], a[n - 1 - i]) + k)))
        r[max(a[i], a[n - 1 - i]) + k] = 1;
      else
        ++r[max(a[i], a[n - 1 - i]) + k];
      if (!(s.count(a[i] + a[n - i - 1])))
        s[a[i] + a[n - i - 1]] = 1;
      else
        ++s[a[i] + a[n - i - 1]];
    }
    ans = n / 2;
    auto li = l.begin(), ri = r.begin(), si = s.begin();
    d = 0;
    for (int i = 0; i < 2 * k + 1; ++i) {
      if (i == (*li).first) {
        d += (*li).second;
        ++li;
      }
      if (i == (*ri).first + 1) {
        d -= (*ri).second;
        ++ri;
      }
      if (i == (*si).first) {
        x = 0 * (*si).second + 1 * (d - (*si).second) + 2 * (n / 2 - d);
        ++si;
      } else
        x = 1 * (d) + 2 * (n / 2 - d);
      if (x < ans) ans = x;
    }
    cout << ans << endl;
  }
}