#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,avx,avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
using namespace std;
void pre() {}
int arr[20000];
int n;
int reverse(int i) {
  for (int j = 0; j < i / 2; j++) swap(arr[j], arr[i - j - 1]);
  return i;
}
int reverse_till(int num) {
  int k = 0;
  for (int j = 0; j < n; j++) {
    if (arr[j] == num) {
      k = j;
      break;
    }
  }
  return reverse(k + 1);
}
int reverse_till_1(int num) {
  int k = 0;
  for (int j = 0; j < n; j++) {
    if (arr[j] == num) {
      k = j;
      break;
    }
  }
  k--;
  return reverse(k + 1);
}
int reverse_till_11(int num) {
  int k = 0;
  for (int j = 0; j < n; j++) {
    if (arr[j] == num) {
      k = j;
      break;
    }
  }
  k++;
  return reverse(k + 1);
}
void print_arr() {}
void solve() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
  }
  vector<int> v;
  for (int i = 0; i < n; i++) {
    if (arr[i] % 2 == i % 2) {
      cout << -1 << "\n";
      return;
    }
  }
  for (int i = 0; n - 2 * i > 1; i += 1) {
    int a = n - 2 * i;
    int b = n - 1 - 2 * i;
    print_arr();
    v.push_back(reverse_till(a));
    print_arr();
    v.push_back(reverse_till_1(b));
    print_arr();
    v.push_back(reverse_till_11(b));
    print_arr();
    v.push_back(reverse(3));
    print_arr();
    v.push_back(reverse(a));
  }
  print_arr();
  cout << v.size() << "\n";
  for (auto i : v) {
    cout << i << " ";
  }
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  pre();
  long long int num = 1;
  cin >> num;
  for (long long int i = 0; i < num; i++) {
    solve();
  }
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,avx,avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
using namespace std;
void pre() {}
int arr[20000];
int n;
int reverse(int i) {
  for (int j = 0; j < i / 2; j++) swap(arr[j], arr[i - j - 1]);
  return i;
}
int reverse_till(int num) {
  int k = 0;
  for (int j = 0; j < n; j++) {
    if (arr[j] == num) {
      k = j;
      break;
    }
  }
  return reverse(k + 1);
}
int reverse_till_1(int num) {
  int k = 0;
  for (int j = 0; j < n; j++) {
    if (arr[j] == num) {
      k = j;
      break;
    }
  }
  k--;
  return reverse(k + 1);
}
int reverse_till_11(int num) {
  int k = 0;
  for (int j = 0; j < n; j++) {
    if (arr[j] == num) {
      k = j;
      break;
    }
  }
  k++;
  return reverse(k + 1);
}
void print_arr() {}
void solve() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
  }
  vector<int> v;
  for (int i = 0; i < n; i++) {
    if (arr[i] % 2 == i % 2) {
      cout << -1 << "\n";
      return;
    }
  }
  for (int i = 0; n - 2 * i > 1; i += 1) {
    int a = n - 2 * i;
    int b = n - 1 - 2 * i;
    print_arr();
    v.push_back(reverse_till(a));
    print_arr();
    v.push_back(reverse_till_1(b));
    print_arr();
    v.push_back(reverse_till_11(b));
    print_arr();
    v.push_back(reverse(3));
    print_arr();
    v.push_back(reverse(a));
  }
  print_arr();
  cout << v.size() << "\n";
  for (auto i : v) {
    cout << i << " ";
  }
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  pre();
  long long int num = 1;
  cin >> num;
  for (long long int i = 0; i < num; i++) {
    solve();
  }
}

```