#include <bits/stdc++.h>
using namespace std;
void rev(vector<int> &a, int l) { reverse(a.begin(), a.begin() + l); }
void swap_h(vector<int> &a, int n, vector<int> &v) {
  int i = (find(a.begin(), a.end(), n) - a.begin());
  rev(a, i + 1);
  v.push_back(i + 1);
  int j = (find(a.begin(), a.end(), n - 1) - a.begin());
  rev(a, j);
  v.push_back(j);
  rev(a, j + 2);
  v.push_back(j + 2);
  rev(a, 3);
  v.push_back(3);
  rev(a, n);
  v.push_back(n);
}
int main() {
  int tt;
  cin >> tt;
  while (tt--) {
    int n;
    cin >> n;
    vector<int> a(n);
    bool pos = true;
    for (int i = 0; i < n; i++) {
      cin >> a[i];
      if (a[i] % 2 == i % 2) pos = false;
    }
    if (!pos) {
      cout << -1 << '\n';
      continue;
    }
    vector<int> v;
    for (int i = n; i >= 3; i -= 2) {
      swap_h(a, i, v);
    }
    cout << v.size() << '\n';
    for (int i = 0; i < v.size(); i++) {
      cout << v[i] << ' ';
    }
    cout << '\n';
  }
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
void rev(vector<int> &a, int l) { reverse(a.begin(), a.begin() + l); }
void swap_h(vector<int> &a, int n, vector<int> &v) {
  int i = (find(a.begin(), a.end(), n) - a.begin());
  rev(a, i + 1);
  v.push_back(i + 1);
  int j = (find(a.begin(), a.end(), n - 1) - a.begin());
  rev(a, j);
  v.push_back(j);
  rev(a, j + 2);
  v.push_back(j + 2);
  rev(a, 3);
  v.push_back(3);
  rev(a, n);
  v.push_back(n);
}
int main() {
  int tt;
  cin >> tt;
  while (tt--) {
    int n;
    cin >> n;
    vector<int> a(n);
    bool pos = true;
    for (int i = 0; i < n; i++) {
      cin >> a[i];
      if (a[i] % 2 == i % 2) pos = false;
    }
    if (!pos) {
      cout << -1 << '\n';
      continue;
    }
    vector<int> v;
    for (int i = n; i >= 3; i -= 2) {
      swap_h(a, i, v);
    }
    cout << v.size() << '\n';
    for (int i = 0; i < v.size(); i++) {
      cout << v[i] << ' ';
    }
    cout << '\n';
  }
}

```