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>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
using namespace std;
const long long MOD = 1e9 + 7;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    long long n, k;
    cin >> n >> k;
    vector<long long> ar(n);
    for (long long i = 0; i < n; i++) cin >> ar[i];
    vector<long long> mi;
    vector<long long> mx;
    vector<long long> h(2 * k + 1, 0);
    for (long long i = 0; i < n / 2; i++) {
      long long x = min(ar[i], ar[n - i - 1]);
      long long y = max(ar[i], ar[n - i - 1]);
      mi.push_back(x);
      mx.push_back(y);
      ++h[x + y];
    }
    sort(mi.begin(), mi.end());
    sort(mx.begin(), mx.end());
    long long m = mi.size();
    long long ans = n;
    for (long long i = 2; i <= k; i++) {
      long long ct = n / 2 - h[i];
      long long pos = lower_bound(mi.begin(), mi.end(), i) - mi.begin();
      if (pos <= m) ct += m - pos;
      ans = min(ans, ct);
    }
    m = mx.size();
    for (long long i = k + 1; i <= 2 * k; i++) {
      long long ct = n / 2 - h[i];
      long long pos = lower_bound(mx.begin(), mx.end(), i - k) - mx.begin();
      if (pos <= m) ct += pos;
      ans = min(ans, ct);
    }
    cout << ans << "\n";
  }
  return 0;
}
