#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using vi = vector<int>;
using pi = pair<int, int>;
const int mxn = 2021 + 5;
int n, a[mxn];
void solve() {
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    --a[i];
  }
  for (int i = 0; i < n; ++i) {
    if ((a[i] & 1) != (i & 1)) {
      cout << "-1\n";
      return;
    }
  }
  vi ans;
  for (int i = n - 1; i >= 2; i -= 2) {
    int x = 0, y = 0;
    while (a[x] != i) ++x;
    ans.push_back(x);
    reverse(a, a + x + 1);
    while (a[y] != i - 1) ++y;
    ans.push_back(y - 1);
    reverse(a, a + y);
    ans.push_back(y + 1);
    reverse(a, a + y + 2);
    ans.push_back(2);
    reverse(a, a + 3);
    ans.push_back(i);
    reverse(a, a + i + 1);
  }
  cout << (int)(ans).size() << '\n';
  for (const int x : ans) cout << x + 1 << ' ';
  cout << '\n';
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) solve();
}

### Prompt

In cpp, your task is to solve 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;
using ll = long long;
using vi = vector<int>;
using pi = pair<int, int>;
const int mxn = 2021 + 5;
int n, a[mxn];
void solve() {
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    --a[i];
  }
  for (int i = 0; i < n; ++i) {
    if ((a[i] & 1) != (i & 1)) {
      cout << "-1\n";
      return;
    }
  }
  vi ans;
  for (int i = n - 1; i >= 2; i -= 2) {
    int x = 0, y = 0;
    while (a[x] != i) ++x;
    ans.push_back(x);
    reverse(a, a + x + 1);
    while (a[y] != i - 1) ++y;
    ans.push_back(y - 1);
    reverse(a, a + y);
    ans.push_back(y + 1);
    reverse(a, a + y + 2);
    ans.push_back(2);
    reverse(a, a + 3);
    ans.push_back(i);
    reverse(a, a + i + 1);
  }
  cout << (int)(ans).size() << '\n';
  for (const int x : ans) cout << x + 1 << ' ';
  cout << '\n';
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) solve();
}

```