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>
template <class T>
inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
constexpr long long MAX = 5100000;
constexpr long long INF = 1LL << 60;
constexpr int inf = 1000000007;
constexpr long long mod = 1000000007LL;
const long double PI = acos((long double)(-1));
using namespace std;
void f(int idx, vector<int>& a) {
  swap(a[idx + 1], a[idx + 2]);
  swap(a[idx], a[idx + 1]);
}
int find(int s, vector<int>& a) {
  for (int i = 0; i < a.size(); i++)
    if (a[i] == s) return i;
}
void solve() {
  int n;
  scanf("%d", &n);
  vector<int> a(n);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]), a[i]--;
  vector<pair<int, int>> c(n);
  for (int i = 0; i < n; i++) c[i] = {a[i], i};
  sort(c.begin(), c.end());
  for (int i = 0; i < n; i++) a[c[i].second] = i;
  vector<int> res;
  for (int s = 0; s < n - 2; s++) {
    int idx = find(s, a);
    while (idx != s) {
      if (idx - 1 == s)
        f(idx - 1, a), res.emplace_back(idx - 1), idx++;
      else
        f(idx - 2, a), res.emplace_back(idx - 2), idx -= 2;
    }
  }
  auto b = a;
  sort(b.begin(), b.end());
  if (a != b) {
    for (int i = 0; i + 1 < n; i++) {
      if (c[i].first == c[i + 1].first) {
        swap(c[i], c[i + 1]);
        break;
      }
    }
    for (int i = 0; i < n; i++) a[c[i].second] = i;
    vector<int> res;
    for (int s = 0; s < n - 2; s++) {
      int idx = find(s, a);
      while (idx != s) {
        if (idx - 1 == s)
          f(idx - 1, a), res.emplace_back(idx - 1), idx++;
        else
          f(idx - 2, a), res.emplace_back(idx - 2), idx -= 2;
      }
    }
    auto b = a;
    sort(b.begin(), b.end());
    if (a != b)
      puts("-1");
    else {
      printf("%d\n", (int)res.size());
      for (int i = 0; i < res.size(); i++)
        printf("%d%c", res[i] + 1, " \n"[i + 1 == res.size()]);
    }
  } else {
    printf("%d\n", (int)res.size());
    for (int i = 0; i < res.size(); i++)
      printf("%d%c", res[i] + 1, " \n"[i + 1 == res.size()]);
  }
}
int main() {
  int kkt;
  scanf("%d", &kkt);
  while (kkt--) {
    solve();
  }
  return 0;
}