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 int INF = 2e9;
const long long ML = 4e18;
void opt(int i, vector<int>& a) {
  int p = a[i];
  int q = a[i + 1];
  a[i] = a[i + 2];
  a[i + 1] = p;
  a[i + 2] = q;
}
bool help(vector<int>& ans, vector<int>& a) {
  int n = a.size();
  for (int i = (0); i < (n - 2); ++i) {
    int p = min_element(a.begin() + i, a.end()) - a.begin();
    if (p == i) continue;
    if ((p - i) % 2) {
      if (p == n - 1) {
        ans.push_back(p - 2);
        opt(p - 2, a);
        p = p - 2;
      }
      ans.push_back(p - 1);
      opt(p - 1, a);
      p++;
    }
    while (p > i) {
      ans.push_back(p - 2);
      opt(p - 2, a);
      p -= 2;
    }
  }
  if (a[n - 1] < a[n - 2]) return false;
  return true;
}
void solve() {
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = (0); i < (n); ++i) cin >> a[i];
  vector<int> ans;
  vector<int> d(n);
  for (int i = (0); i < (n); ++i) d[i] = i;
  auto f = [&a](int i, int j) { return a[i] < a[j]; };
  sort((d).begin(), (d).end(), f);
  vector<int> idx(n);
  for (int i = (0); i < (n); ++i) idx[d[i]] = i;
  if (help(ans, idx)) {
    cout << ans.size() << endl;
    for (int i : ans) cout << i + 1 << " ";
    cout << '\n';
    return;
  }
  map<int, int> mp;
  int x, y;
  bool ok = false;
  for (int i = (0); i < (n); ++i) {
    if (!mp.count(a[i]))
      mp[a[i]] = i;
    else {
      x = mp[a[i]];
      y = i;
      ok = true;
      break;
    }
  }
  if (ok) {
    for (int i = (0); i < (n); ++i) idx[d[i]] = i;
    swap(idx[x], idx[y]);
    ans.clear();
    if (help(ans, idx)) {
      cout << ans.size() << endl;
      for (int i : ans) cout << i + 1 << " ";
      cout << '\n';
      return;
    }
  }
  cout << -1 << endl;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int T;
  cin >> T;
  for (int kase = 1; kase <= T; ++kase) {
    solve();
  }
  return 0;
}