#include <bits/stdc++.h>
using namespace std;
const double pi = 3.14159265358979323846, eps = 1e-9;
const long long M = 1e6, I = 1e17 + 5;
const int mxn = 2100, mxn1 = 2000, lg = 20;
int t, n;
vector<int> arr, ans;
void out() {
  for (int i : arr) cout << i << " ";
  cout << endl;
}
void solve() {
  arr.clear();
  ans.clear();
  cin >> n;
  bool f = 0;
  for (int i = 0; i < n; i++) {
    int k;
    cin >> k;
    arr.push_back(k);
    if (i % 2 == arr[i] % 2) {
      f = 1;
    }
  }
  if (f) {
    cout << -1 << endl;
    return;
  }
  for (int i = n; i >= 3; i -= 2) {
    int ind = -1;
    for (int j = 0; j < i; j++) {
      if (arr[j] == i) {
        ind = j;
        break;
      }
    }
    ans.push_back(ind + 1);
    reverse(arr.begin(), arr.begin() + ind + 1);
    for (int j = 0; j < i; j++) {
      if (arr[j] == i - 1) {
        ind = j;
        break;
      }
    }
    ans.push_back(ind);
    reverse(arr.begin(), arr.begin() + ind);
    ans.push_back(ind + 2);
    reverse(arr.begin(), arr.begin() + ind + 2);
    ans.push_back(3);
    reverse(arr.begin(), arr.begin() + 3);
    ans.push_back(i);
    reverse(arr.begin(), arr.begin() + i);
  }
  cout << ans.size() << endl;
  for (int i : ans) cout << i << " ";
  cout << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  cin >> t;
  while (t--) solve();
}

### 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;
const double pi = 3.14159265358979323846, eps = 1e-9;
const long long M = 1e6, I = 1e17 + 5;
const int mxn = 2100, mxn1 = 2000, lg = 20;
int t, n;
vector<int> arr, ans;
void out() {
  for (int i : arr) cout << i << " ";
  cout << endl;
}
void solve() {
  arr.clear();
  ans.clear();
  cin >> n;
  bool f = 0;
  for (int i = 0; i < n; i++) {
    int k;
    cin >> k;
    arr.push_back(k);
    if (i % 2 == arr[i] % 2) {
      f = 1;
    }
  }
  if (f) {
    cout << -1 << endl;
    return;
  }
  for (int i = n; i >= 3; i -= 2) {
    int ind = -1;
    for (int j = 0; j < i; j++) {
      if (arr[j] == i) {
        ind = j;
        break;
      }
    }
    ans.push_back(ind + 1);
    reverse(arr.begin(), arr.begin() + ind + 1);
    for (int j = 0; j < i; j++) {
      if (arr[j] == i - 1) {
        ind = j;
        break;
      }
    }
    ans.push_back(ind);
    reverse(arr.begin(), arr.begin() + ind);
    ans.push_back(ind + 2);
    reverse(arr.begin(), arr.begin() + ind + 2);
    ans.push_back(3);
    reverse(arr.begin(), arr.begin() + 3);
    ans.push_back(i);
    reverse(arr.begin(), arr.begin() + i);
  }
  cout << ans.size() << endl;
  for (int i : ans) cout << i << " ";
  cout << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  cin >> t;
  while (t--) solve();
}

```