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 main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  for (int dp = 0; dp < t; dp++) {
    int n, k;
    cin >> n >> k;
    vector<int> vec(n);
    for (int i = 0; i < n; i++) cin >> vec[i];
    map<int, int> mymap;
    int i = 0, j = n - 1;
    vector<int> a;
    vector<int> b;
    while (i < j) {
      mymap[vec[i] + vec[j]] += 1;
      int x = min(vec[i], vec[j]);
      int y = max(vec[i], vec[j]);
      b.push_back(1 + x);
      a.push_back(k + y);
      i++;
      j--;
    }
    int ans = pow(10, 15);
    int sum = 0;
    map<int, int>::iterator it = mymap.begin();
    for (; it != mymap.end(); it++) sum += it->second;
    sort(a.begin(), a.end());
    sort(b.begin(), b.end());
    it = mymap.begin();
    for (int i = 1; i <= 2 * k; i++) {
      int val = i;
      int r = mymap[i];
      int cur = sum - r;
      int x = lower_bound(a.begin(), a.end(), val) - a.begin();
      cur += x;
      x = upper_bound(b.begin(), b.end(), val) - b.begin();
      cur += n / 2 - x;
      ans = min(ans, cur);
    }
    cout << ans << endl;
  }
  return 0;
}