#include <bits/stdc++.h>
using namespace std;
void apply(vector<int>& P, int v, vector<int>& moves) {
  moves.push_back(v);
  for (int i = 0; i * 2 < v - 1; ++i) {
    swap(P[i], P[v - 1 - i]);
  }
}
void one() {
  int N;
  cin >> N;
  vector<int> P(N);
  for (int i = 0; i < N; ++i) {
    cin >> P[i];
  }
  vector<int> moves;
  for (int j = N; j >= 3; j -= 2) {
    int index = -1;
    for (int i = 0; i < j; ++i) {
      if (P[i] == j) index = i;
    }
    if (index % 2 != 0) {
      cout << "-1\n";
      return;
    }
    apply(P, index + 1, moves);
    for (int i = 0; i < j; ++i) {
      if (P[i] == j - 1) index = i;
    }
    if (index % 2 != 1) {
      cout << "-1\n";
      return;
    }
    apply(P, index, moves);
    apply(P, index + 2, moves);
    apply(P, 3, moves);
    apply(P, j, moves);
  }
  cout << moves.size() << '\n';
  for (int i = 0; i < moves.size(); ++i)
    cout << moves[i] << " \n"[i == moves.size() - 1];
}
int main() {
  std::ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int TC;
  cin >> TC;
  while (TC-- > 0) {
    one();
  }
  cout << flush;
}

### Prompt

Generate a CPP solution to the following problem:

You have a permutation: an array a = [a_1, a_2, …, a_n] of distinct integers from 1 to n. The length of the permutation n is odd.

You need to sort the permutation in increasing order.

In one step, you can choose any prefix of the permutation with an odd length and reverse it. Formally, if a = [a_1, a_2, …, a_n], you can choose any odd integer p between 1 and n, inclusive, and set a to [a_p, a_{p-1}, …, a_1, a_{p+1}, a_{p+2}, …, a_n].

Find a way to sort a using no more than 5n/2 reversals of the above kind, or determine that such a way doesn't exist. The number of reversals doesn't have to be minimized.

Input

Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.

The first line of each test case contains a single integer n (3 ≤ n ≤ 2021; n is odd) — the length of the permutation.

The second line contains n distinct integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n) — the permutation itself. 

It is guaranteed that the sum of n over all test cases does not exceed 2021.

Output

For each test case, if it's impossible to sort the given permutation in at most 5n/2 reversals, print a single integer -1.

Otherwise, print an integer m (0 ≤ m ≤ 5n/2), denoting the number of reversals in your sequence of steps, followed by m integers p_i (1 ≤ p_i ≤ n; p_i is odd), denoting the lengths of the prefixes of a to be reversed, in chronological order.

Note that m doesn't have to be minimized. If there are multiple answers, print any.

Example

Input


3
3
1 2 3
5
3 4 5 2 1
3
2 1 3


Output


4
3 3 3 3
2
3 5
-1

Note

In the first test case, the permutation is already sorted. Any even number of reversals of the length 3 prefix doesn't change that fact.

In the second test case, after reversing the prefix of length 3 the permutation will change to [5, 4, 3, 2, 1], and then after reversing the prefix of length 5 the permutation will change to [1, 2, 3, 4, 5].

In the third test case, it's impossible to sort the permutation.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
void apply(vector<int>& P, int v, vector<int>& moves) {
  moves.push_back(v);
  for (int i = 0; i * 2 < v - 1; ++i) {
    swap(P[i], P[v - 1 - i]);
  }
}
void one() {
  int N;
  cin >> N;
  vector<int> P(N);
  for (int i = 0; i < N; ++i) {
    cin >> P[i];
  }
  vector<int> moves;
  for (int j = N; j >= 3; j -= 2) {
    int index = -1;
    for (int i = 0; i < j; ++i) {
      if (P[i] == j) index = i;
    }
    if (index % 2 != 0) {
      cout << "-1\n";
      return;
    }
    apply(P, index + 1, moves);
    for (int i = 0; i < j; ++i) {
      if (P[i] == j - 1) index = i;
    }
    if (index % 2 != 1) {
      cout << "-1\n";
      return;
    }
    apply(P, index, moves);
    apply(P, index + 2, moves);
    apply(P, 3, moves);
    apply(P, j, moves);
  }
  cout << moves.size() << '\n';
  for (int i = 0; i < moves.size(); ++i)
    cout << moves[i] << " \n"[i == moves.size() - 1];
}
int main() {
  std::ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int TC;
  cin >> TC;
  while (TC-- > 0) {
    one();
  }
  cout << flush;
}

```