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;
const long double PI = acos(-1);
long long power(long long b, long long e, long long m) {
  if (e == 0) return 1;
  if (e & 1) return b * power(b * b % m, e / 2, m) % m;
  return power(b * b % m, e / 2, m);
}
long long power(long long b, long long e) {
  if (e == 0) return 1;
  if (e & 1) return b * power(b * b, e / 2);
  return power(b * b, e / 2);
}
template <typename T, typename U>
static inline void amin(T &x, U y) {
  if (y < x) x = y;
}
template <typename T, typename U>
static inline void amax(T &x, U y) {
  if (x < y) x = y;
}
template <typename T, typename U>
ostream &operator<<(ostream &os, const pair<T, U> &p) {
  return os << '(' << p.first << "," << p.second << ')';
}
void shift(vector<int> &a, int idx) {
  swap(a[idx + 2], a[idx + 1]);
  swap(a[idx + 1], a[idx]);
}
int _runtimeTerror_() {
  long long n;
  cin >> n;
  vector<int> a(n + 1), b;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  b = a;
  sort(b.begin(), b.end());
  vector<int> ans;
  for (int i = 1; i <= n - 2; ++i) {
    if (a[i] == b[i]) continue;
    for (int j = i + 1; j <= n; ++j) {
      if (b[i] == a[j]) {
        int cur = j - 2;
        for (int k = cur; k >= i; k -= 2) {
          ans.push_back(k);
          shift(a, k);
        }
        if ((j - i) % 2 == 1) {
          ans.push_back(i);
          ans.push_back(i);
          shift(a, i);
          shift(a, i);
        }
      }
    }
  }
  if (a != b && a[n - 2] == a[n]) ans.push_back(n - 2), shift(a, n - 2);
  if (a != b) {
    int idx = -1;
    for (int i = 1; i <= n - 2; ++i) {
      if (a[i] == a[i - 1]) {
        idx = i;
        break;
      }
    }
    if (idx == -1) {
      cout << "-1\n";
      return 0;
    }
    for (int i = n - 2; i >= idx - 1; --i) ans.push_back(i);
  }
  cout << (long long)ans.size() << '\n';
  for (auto k : ans) cout << k << ' ';
  cout << '\n';
  return 0;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int TESTS = 1;
  cin >> TESTS;
  while (TESTS--) _runtimeTerror_();
  return 0;
}