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 Arg1>
void __f(const char* name, Arg1&& arg1) {
  std::cerr << name << " : " << arg1 << '\n';
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  std::cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
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;
}
long long max(long long a, long long b) { return (a > b) ? a : b; }
long long min(long long a, long long b) { return (a < b) ? a : b; }
long long change(long long a[], long long k) {
  long long z = a[k + 2];
  a[k + 2] = a[k + 1];
  a[k + 1] = a[k];
  a[k] = z;
}
vector<long long> v;
void hola(long long a[], long long n, long long so[]) {
  if (n < 3) return;
  bool z = 1;
  for (long long i = 0; i < n; i++)
    if (a[i] != so[i]) z = 0;
  if (z) return;
  change(a, n - 3);
  v.push_back(n - 3);
  if (a[n - 3] == a[n - 1]) {
    change(a, n - 3);
    v.push_back(n - 3);
    return;
  }
  hola(a, n - 1, so);
}
long long solve() {
  v.clear();
  long long n;
  cin >> n;
  long long a[n];
  for (long long i = 0; i < n; ++i) cin >> a[i];
  map<long long, long long> m;
  for (long long i = 0; i < n; i++) m[a[i]];
  long long ind = 0;
  for (auto& i : m) i.second = ind++;
  for (long long i = 0; i < n; i++) a[i] = m[a[i]];
  long long so[n];
  for (long long i = 0; i < n; i++) so[i] = a[i];
  sort(so, so + n);
  for (long long i = 0; i < n - 2; i++) {
    if (a[i] == so[i]) continue;
    long long find = -1;
    for (long long j = i + 1; j < n; j++) {
      if (a[j] == so[i]) {
        find = j;
        break;
      }
    }
    while (find > i + 1) {
      change(a, find - 2);
      v.push_back(find - 2);
      find -= 2;
    }
    if (a[i] == so[i]) continue;
    change(a, i);
    change(a, i);
    v.push_back(i);
    v.push_back(i);
  }
  if (a[n - 1] == n - 1) {
    cout << v.size() << '\n';
    for (auto i : v) cout << i + 1 << " ";
    cout << '\n';
    return 0;
  }
  m.clear();
  for (long long i = 0; i < n; i++) m[a[i]]++;
  bool z = 0;
  for (auto i : m)
    if (i.second >= 2) z = 1;
  if (!z) return cout << -1 << '\n', 0;
  hola(a, n, so);
  if (v.size() <= n * n) {
    cout << v.size() << '\n';
    for (auto i : v) cout << i + 1 << " ";
    cout << '\n';
    return 0;
  } else
    cout << -1 << '\n';
  return 0;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
