#include <bits/stdc++.h>
using namespace std;
int n, a[2040];
vector<int> res;
int find(int x) {
  for (int i = 1; i <= n; i++)
    if (a[i] == x) return i;
}
void add(int x) {
  res.push_back(x);
  for (int i = 1, j = x; i < j; i++, j--) swap(a[i], a[j]);
}
void solve() {
  res.clear();
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= n; i++)
    if (i % 2 != a[i] % 2) {
      puts("-1");
      return;
    }
  for (int i = n; i > 1; i -= 2) {
    int posA = find(i);
    add(posA);
    int posB = find(i - 1);
    add(posB - 1);
    add(posB + 1);
    add(3);
    add(i);
  }
  cout << res.size() << "\n";
  for (auto x : res) cout << x << " ";
  puts("");
}
int main() {
  int Tc;
  for (cin >> Tc; Tc--; solve())
    ;
}

### Prompt

Please create a solution in cpp 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;
int n, a[2040];
vector<int> res;
int find(int x) {
  for (int i = 1; i <= n; i++)
    if (a[i] == x) return i;
}
void add(int x) {
  res.push_back(x);
  for (int i = 1, j = x; i < j; i++, j--) swap(a[i], a[j]);
}
void solve() {
  res.clear();
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= n; i++)
    if (i % 2 != a[i] % 2) {
      puts("-1");
      return;
    }
  for (int i = n; i > 1; i -= 2) {
    int posA = find(i);
    add(posA);
    int posB = find(i - 1);
    add(posB - 1);
    add(posB + 1);
    add(3);
    add(i);
  }
  cout << res.size() << "\n";
  for (auto x : res) cout << x << " ";
  puts("");
}
int main() {
  int Tc;
  for (cin >> Tc; Tc--; solve())
    ;
}

```