#include <bits/stdc++.h>
using namespace std;
int A[3003];
vector<int> F;
void solve(int x, int y, int n) {
  int num = A[x];
  F.push_back(y);
  reverse(A + 1, A + 1 + y);
  if (x <= y) {
    for (int i = 1; i <= n; i++)
      if (A[i] == num) {
        x = i;
        break;
      }
  }
  F.push_back(x - 1);
  reverse(A + 1, A + x);
  F.push_back(x + 1);
  reverse(A + 1, A + x + 2);
  F.push_back(3);
  reverse(A + 1, A + 4);
  F.push_back(n);
  reverse(A + 1, A + 1 + n);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL), cout.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    for (int i = 1; i <= n; i++) cin >> A[i];
    int flag = 0;
    for (int i = 1; i <= n; i++)
      if (A[i] % 2 != i % 2) flag = 1;
    if (flag == 1) {
      cout << -1 << "\n";
      continue;
    }
    F.clear();
    for (int i = n; i >= 3; i -= 2) {
      int x = -1, y = -1;
      for (int j = 1; j <= i; j++) {
        if (A[j] == i - 1) x = j;
        if (A[j] == i) y = j;
      }
      solve(x, y, i);
    }
    cout << F.size() << "\n";
    for (auto f : F) cout << f << " ";
    cout << "\n";
  }
  return 0;
}

### 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;
int A[3003];
vector<int> F;
void solve(int x, int y, int n) {
  int num = A[x];
  F.push_back(y);
  reverse(A + 1, A + 1 + y);
  if (x <= y) {
    for (int i = 1; i <= n; i++)
      if (A[i] == num) {
        x = i;
        break;
      }
  }
  F.push_back(x - 1);
  reverse(A + 1, A + x);
  F.push_back(x + 1);
  reverse(A + 1, A + x + 2);
  F.push_back(3);
  reverse(A + 1, A + 4);
  F.push_back(n);
  reverse(A + 1, A + 1 + n);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL), cout.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    for (int i = 1; i <= n; i++) cin >> A[i];
    int flag = 0;
    for (int i = 1; i <= n; i++)
      if (A[i] % 2 != i % 2) flag = 1;
    if (flag == 1) {
      cout << -1 << "\n";
      continue;
    }
    F.clear();
    for (int i = n; i >= 3; i -= 2) {
      int x = -1, y = -1;
      for (int j = 1; j <= i; j++) {
        if (A[j] == i - 1) x = j;
        if (A[j] == i) y = j;
      }
      solve(x, y, i);
    }
    cout << F.size() << "\n";
    for (auto f : F) cout << f << " ";
    cout << "\n";
  }
  return 0;
}

```