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;
int n;
int A[501];
void add(int i) {
  swap(A[i], A[i + 1]);
  swap(A[i], A[i + 2]);
}
void solve() {
  cin >> n;
  int cnt = 0, same = 0, a, b;
  for (int i = 1; i <= n; ++i) {
    cin >> A[i];
    for (int j = 1; j < i; ++j) {
      if (A[j] > A[i]) ++cnt;
      if (A[i] == A[j]) same = 1, a = i, b = j;
    }
  }
  if (!same && cnt % 2 == 1) {
    printf("-1\n");
    return;
  }
  vector<pair<int, int> > P;
  for (int i = 1; i <= n; ++i) {
    P.emplace_back(A[i], i);
  }
  sort(P.begin(), P.end());
  for (int i = 1; i <= n; ++i) {
    A[P[i - 1].second] = i;
  }
  cnt = 0;
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j < i; ++j) {
      if (A[j] > A[i]) ++cnt;
    }
  }
  if (cnt & 1) swap(A[a], A[b]);
  vector<int> ans;
  for (int v = n; v >= 1; --v) {
    int p;
    for (int i = 1; i <= v; ++i)
      if (A[i] == v) p = i;
    for (; p < v; ++p) {
      int i = max(1, min(p, v - 2));
      add(i);
      ans.push_back(i);
    }
  }
  printf("%u\n", ans.size());
  for (int i : ans) printf("%d ", i);
  printf("\n");
  for (int i = 1; i <= n; ++i)
    ;
  ;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.precision(30);
  cout << fixed;
  int T;
  cin >> T;
  while (T--) solve();
  return 0;
}
