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;
template <typename A, typename B>
ostream& operator<<(ostream& os, const pair<A, B>& p) {
  os << "(" << p.first << "," << p.second << ") ";
  return os;
}
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& v) {
  for (auto i : v) os << i << " ";
  return os;
}
const long long int mxN = 3e5;
const long long int mxK = 2e5;
const long long int mod = 1e9 + 7;
long long int dp[2 * mxK + 1];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long int t;
  cin >> t;
  while (t--) {
    long long int n, k;
    cin >> n >> k;
    long long int a[n];
    for (long long int i = 0; i < n; i++) cin >> a[i];
    for (int i = 0; i <= 2 * k; i++) dp[i] = 0;
    for (long long int i = 0; i < n / 2; i++) {
      long long int s = a[i] + a[n - i - 1];
      long long int l = s - max(a[i] - 1, a[n - i - 1] - 1);
      long long int r = s + max(k - a[i], k - a[n - i - 1]);
      dp[0] += 2;
      dp[max(0ll, l)]--;
      dp[s]--;
      dp[s + 1]++;
      dp[r + 1]++;
    }
    long long int mn = 1e9;
    for (int i = 1; i <= 2 * k; i++) {
      dp[i] += dp[i - 1];
      mn = min(dp[i], mn);
    }
    cout << mn << "\n";
  }
}
