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 ll = long long;
using ld = long double;
using namespace std;
template <class T>
bool chmax(T& a, const T& b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool chmin(T& a, const T& b) {
  if (b < a) {
    a = b;
    return 1;
  }
  return 0;
}
ll mod = 1e9 + 7;
ll MOD = 1e9 + 7;
ll INF = 1e18;
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout << fixed << setprecision(10);
  ll t;
  cin >> t;
  for (int index = 0; index < (int)(t); index++) {
    ll n, k;
    cin >> n >> k;
    vector<ll> vec(n);
    for (int i = 0; i < (int)(n); i++) cin >> vec[i];
    vector<ll> reg(2 * k + 10, 0);
    for (int i = 0; i < (int)(n / 2); i++) {
      ll val1 = i;
      ll val2 = (n - i - 1);
      reg[0] += 2;
      reg[min(vec[val1], vec[val2]) + 1]--;
      reg[vec[val1] + vec[val2]]--;
      reg[vec[val1] + vec[val2] + 1]++;
      reg[max(vec[val1], vec[val2]) + k + 1]++;
    }
    for (ll(i) = (1); (i) < (2 * k + 3); ++(i)) {
      reg[i] += reg[i - 1];
    }
    ll res = INF;
    for (ll(i) = (0); (i) < (2 * k + 1); ++(i)) {
      chmin(res, reg[i]);
    }
    cout << res << endl;
  }
}
