#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
void printn(vector<int> nums) {
  for (int a : nums) {
    cout << a << " ";
  }
  cout << '\n';
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t, n;
  cin >> t;
  for (int test = 0; test < t; test++) {
    cin >> n;
    vector<int> nums(n);
    bool Pos = true;
    for (int a = 0; a < n; a++) {
      cin >> nums[a];
      if ((nums[a] + 1 - a) % 2 != 0) {
        Pos = false;
      }
    }
    if (!Pos) {
      cout << -1 << '\n';
      continue;
    }
    int pindex;
    int oindex;
    int act = n;
    cout << 5 * (n - 1) / 2 << '\n';
    while (act > 1) {
      for (int a = 0; a < n; a++) {
        if (nums[a] == act) {
          oindex = a;
        }
        if (nums[a] == act - 1) {
          pindex = a;
        }
      }
      cout << oindex + 1 << " ";
      reverse(nums.begin(), nums.begin() + oindex + 1);
      if (oindex > pindex) {
        pindex = oindex - pindex;
      }
      oindex = 0;
      cout << pindex << " ";
      reverse(nums.begin(), nums.begin() + pindex);
      cout << pindex + 2 << " ";
      reverse(nums.begin(), nums.begin() + pindex + 2);
      cout << 3 << " ";
      reverse(nums.begin(), nums.begin() + 3);
      cout << act << " ";
      reverse(nums.begin(), nums.begin() + act);
      act -= 2;
    }
    cout << '\n';
  }
  cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
  return 0;
}

### Prompt

Create a solution in Cpp for 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>
#pragma GCC optimize("Ofast")
using namespace std;
void printn(vector<int> nums) {
  for (int a : nums) {
    cout << a << " ";
  }
  cout << '\n';
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t, n;
  cin >> t;
  for (int test = 0; test < t; test++) {
    cin >> n;
    vector<int> nums(n);
    bool Pos = true;
    for (int a = 0; a < n; a++) {
      cin >> nums[a];
      if ((nums[a] + 1 - a) % 2 != 0) {
        Pos = false;
      }
    }
    if (!Pos) {
      cout << -1 << '\n';
      continue;
    }
    int pindex;
    int oindex;
    int act = n;
    cout << 5 * (n - 1) / 2 << '\n';
    while (act > 1) {
      for (int a = 0; a < n; a++) {
        if (nums[a] == act) {
          oindex = a;
        }
        if (nums[a] == act - 1) {
          pindex = a;
        }
      }
      cout << oindex + 1 << " ";
      reverse(nums.begin(), nums.begin() + oindex + 1);
      if (oindex > pindex) {
        pindex = oindex - pindex;
      }
      oindex = 0;
      cout << pindex << " ";
      reverse(nums.begin(), nums.begin() + pindex);
      cout << pindex + 2 << " ";
      reverse(nums.begin(), nums.begin() + pindex + 2);
      cout << 3 << " ";
      reverse(nums.begin(), nums.begin() + 3);
      cout << act << " ";
      reverse(nums.begin(), nums.begin() + act);
      act -= 2;
    }
    cout << '\n';
  }
  cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
  return 0;
}

```