#include <bits/stdc++.h>
using namespace std;
long long a[2500];
vector<long long> answer;
long long find(long long v) {
  for (register long long i = 1;; i++)
    if (a[i] == v) return i;
  return -1;
}
void reverse(long long p) {
  if (p == 1) return;
  reverse(a + 1, a + p + 1);
  answer.push_back(p);
  return;
}
void work(void) {
  long long n;
  cin >> n;
  for (register long long i = 1; i <= n; i++) cin >> a[i];
  for (register long long i = 1; i <= n; i++)
    if (a[i] % 2 != i % 2) return cout << -1 << endl, void();
  answer.clear();
  for (register long long i = n; i > 1; i -= 2) {
    if (a[i] == i && a[i - 1] == i - 1) continue;
    long long p = find(i);
    reverse(p);
    p = find(i - 1);
    reverse(p - 1);
    reverse(p + 1);
    reverse(3);
    reverse(i);
  }
  cout << answer.size() << endl;
  for (vector<long long>::iterator i = answer.begin(); i != answer.end(); i++)
    cout << *i << ' ';
  cout << endl;
  return;
}
signed main() {
  ios::sync_with_stdio(false);
  long long _ = 1;
  cin >> _;
  while (_--) work();
  return 0;
}

### Prompt

Your task is to create 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;
long long a[2500];
vector<long long> answer;
long long find(long long v) {
  for (register long long i = 1;; i++)
    if (a[i] == v) return i;
  return -1;
}
void reverse(long long p) {
  if (p == 1) return;
  reverse(a + 1, a + p + 1);
  answer.push_back(p);
  return;
}
void work(void) {
  long long n;
  cin >> n;
  for (register long long i = 1; i <= n; i++) cin >> a[i];
  for (register long long i = 1; i <= n; i++)
    if (a[i] % 2 != i % 2) return cout << -1 << endl, void();
  answer.clear();
  for (register long long i = n; i > 1; i -= 2) {
    if (a[i] == i && a[i - 1] == i - 1) continue;
    long long p = find(i);
    reverse(p);
    p = find(i - 1);
    reverse(p - 1);
    reverse(p + 1);
    reverse(3);
    reverse(i);
  }
  cout << answer.size() << endl;
  for (vector<long long>::iterator i = answer.begin(); i != answer.end(); i++)
    cout << *i << ' ';
  cout << endl;
  return;
}
signed main() {
  ios::sync_with_stdio(false);
  long long _ = 1;
  cin >> _;
  while (_--) work();
  return 0;
}

```