#include <bits/stdc++.h>
using namespace std;
template <class T>
ostream& operator<<(ostream& out, vector<T>& v) {
  out << "[";
  for (auto k : v) out << k << " ";
  out << "]"
      << "\n";
  return out;
}
template <class T>
ostream& operator<<(ostream& out, set<T> s) {
  out << "{";
  for (auto k : s) out << k << " ";
  out << "}"
      << "\n";
  return out;
}
template <class T, class U>
ostream& operator<<(ostream& out, pair<T, U> p) {
  out << "[ " << p.first << " , " << p.second << " ] ";
  return out;
}
template <class T, class U>
istream& operator>>(istream& in, pair<T, U>& p) {
  in >> p.first >> p.second;
  return in;
}
void Caso() {
  int n;
  cin >> n;
  vector<int> in(n);
  for (int i = 0; i < int(n); i++) cin >> in[i];
  vector<int> res;
  int np = n - (1 - n % 2);
  for (int i = 0; i < int(n); i++)
    if (in[i] % 2 == i % 2) {
      cout << -1 << '\n';
      return;
    }
  while (np > 1) {
    int y = np - 1;
    int y2 = 0;
    while (in[y2] != y) y2++;
    int x = 0;
    while (in[x] != np) x++;
    if (y2 == n - 1) {
      cout << -1 << '\n';
      return;
    }
    if (y2 != x - 1) {
      res.push_back(x + 1);
      reverse(in.begin(), in.begin() + x + 1);
      ;
      if (y2 < x) y2 = x - y2;
      x = 0;
      res.push_back(y2);
      reverse(in.begin(), in.begin() + y2);
      ;
      x = y2 - 1;
      res.push_back(y2 + 2);
      reverse(in.begin(), in.begin() + y2 + 2);
      ;
      x = 2;
      y2 = 1;
    }
    if (x + 1 == np) {
      np -= 2;
      continue;
    }
    if (x % 2 == 1) {
      cout << -1 << '\n';
      return;
    }
    if (x) res.push_back(x + 1);
    reverse(in.begin(), in.begin() + x + 1);
    ;
    res.push_back(np);
    reverse(in.begin(), in.begin() + np);
    ;
    np -= 2;
  }
  for (int i = 0; i < int(n - 1); i++)
    if (in[i] > in[i + 1]) {
      cout << -1 << '\n';
      return;
    }
  cout << res.size() << '\n';
  for (auto x : res) cout << x << " ";
  cout << '\n';
}
int main() {
  cin.tie(0);
  cin.sync_with_stdio(0);
  int t;
  cin >> t;
  while (t--) Caso();
  return 0;
}

### Prompt

Develop a solution in CPP 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>
using namespace std;
template <class T>
ostream& operator<<(ostream& out, vector<T>& v) {
  out << "[";
  for (auto k : v) out << k << " ";
  out << "]"
      << "\n";
  return out;
}
template <class T>
ostream& operator<<(ostream& out, set<T> s) {
  out << "{";
  for (auto k : s) out << k << " ";
  out << "}"
      << "\n";
  return out;
}
template <class T, class U>
ostream& operator<<(ostream& out, pair<T, U> p) {
  out << "[ " << p.first << " , " << p.second << " ] ";
  return out;
}
template <class T, class U>
istream& operator>>(istream& in, pair<T, U>& p) {
  in >> p.first >> p.second;
  return in;
}
void Caso() {
  int n;
  cin >> n;
  vector<int> in(n);
  for (int i = 0; i < int(n); i++) cin >> in[i];
  vector<int> res;
  int np = n - (1 - n % 2);
  for (int i = 0; i < int(n); i++)
    if (in[i] % 2 == i % 2) {
      cout << -1 << '\n';
      return;
    }
  while (np > 1) {
    int y = np - 1;
    int y2 = 0;
    while (in[y2] != y) y2++;
    int x = 0;
    while (in[x] != np) x++;
    if (y2 == n - 1) {
      cout << -1 << '\n';
      return;
    }
    if (y2 != x - 1) {
      res.push_back(x + 1);
      reverse(in.begin(), in.begin() + x + 1);
      ;
      if (y2 < x) y2 = x - y2;
      x = 0;
      res.push_back(y2);
      reverse(in.begin(), in.begin() + y2);
      ;
      x = y2 - 1;
      res.push_back(y2 + 2);
      reverse(in.begin(), in.begin() + y2 + 2);
      ;
      x = 2;
      y2 = 1;
    }
    if (x + 1 == np) {
      np -= 2;
      continue;
    }
    if (x % 2 == 1) {
      cout << -1 << '\n';
      return;
    }
    if (x) res.push_back(x + 1);
    reverse(in.begin(), in.begin() + x + 1);
    ;
    res.push_back(np);
    reverse(in.begin(), in.begin() + np);
    ;
    np -= 2;
  }
  for (int i = 0; i < int(n - 1); i++)
    if (in[i] > in[i + 1]) {
      cout << -1 << '\n';
      return;
    }
  cout << res.size() << '\n';
  for (auto x : res) cout << x << " ";
  cout << '\n';
}
int main() {
  cin.tie(0);
  cin.sync_with_stdio(0);
  int t;
  cin >> t;
  while (t--) Caso();
  return 0;
}

```