#include <bits/stdc++.h>
using namespace std;
const int SEED = chrono::steady_clock::now().time_since_epoch().count();
mt19937 rng(SEED);
void test_case() {
  int n;
  cin >> n;
  bool ok = true;
  vector<int> v(n);
  for (auto &x : v) cin >> x;
  for (int i = 0; i < n; i += 2)
    if (v[i] % 2 == 0) ok = false;
  if (ok) {
    vector<int> op;
    for (int i = n; i > 1; i -= 2) {
      int a, b;
      auto getPos = [&](int x) {
        for (int j = 0; j < n; j++)
          if (v[j] == x) return j;
        return -1;
      };
      a = getPos(i);
      if (a != 0) {
        op.push_back(a);
        reverse(v.begin(), v.begin() + a + 1);
      }
      b = getPos(i - 1);
      op.push_back(b - 1);
      reverse(v.begin(), v.begin() + b);
      op.push_back(b + 1);
      reverse(v.begin(), v.begin() + b + 2);
      op.push_back(2);
      reverse(v.begin(), v.begin() + 3);
      op.push_back(i - 1);
      reverse(v.begin(), v.begin() + i);
    }
    cout << op.size() << endl;
    for (auto x : op) cout << x + 1 << " ";
    cout << endl;
  } else
    cout << -1 << endl;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t = 1;
  cin >> t;
  for (int i = 1; i <= t; i++) {
    test_case();
  }
  return 0;
}

### Prompt

Your challenge is to write 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 int SEED = chrono::steady_clock::now().time_since_epoch().count();
mt19937 rng(SEED);
void test_case() {
  int n;
  cin >> n;
  bool ok = true;
  vector<int> v(n);
  for (auto &x : v) cin >> x;
  for (int i = 0; i < n; i += 2)
    if (v[i] % 2 == 0) ok = false;
  if (ok) {
    vector<int> op;
    for (int i = n; i > 1; i -= 2) {
      int a, b;
      auto getPos = [&](int x) {
        for (int j = 0; j < n; j++)
          if (v[j] == x) return j;
        return -1;
      };
      a = getPos(i);
      if (a != 0) {
        op.push_back(a);
        reverse(v.begin(), v.begin() + a + 1);
      }
      b = getPos(i - 1);
      op.push_back(b - 1);
      reverse(v.begin(), v.begin() + b);
      op.push_back(b + 1);
      reverse(v.begin(), v.begin() + b + 2);
      op.push_back(2);
      reverse(v.begin(), v.begin() + 3);
      op.push_back(i - 1);
      reverse(v.begin(), v.begin() + i);
    }
    cout << op.size() << endl;
    for (auto x : op) cout << x + 1 << " ";
    cout << endl;
  } else
    cout << -1 << endl;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t = 1;
  cin >> t;
  for (int i = 1; i <= t; i++) {
    test_case();
  }
  return 0;
}

```