#include <bits/stdc++.h>
using namespace std;
const long long mod = 1000000007, inf = 2e18;
const int N = 200001;
const double ep = 1e-10;
void Darling(int kase) {
  int n;
  cin >> n;
  int a[n];
  for (int i = 0; (1 < 0 && i > n) || (1 > 0 && i < n); i += 1) cin >> a[i];
  for (int i = 0; (1 < 0 && i > n) || (1 > 0 && i < n); i += 1)
    if (a[i] % 2 == i % 2) {
      cout << -1 << '\n';
      return;
    }
  vector<int> z;
  int o = 1;
  while (o < n) {
    int i1 = 0, i2 = 1;
    for (int i = 0; (1 < 0 && i > n) || (1 > 0 && i < n); i += 1) {
      if (a[i] == o) i1 = i;
      if (a[i] == o + 1) i2 = i;
    }
    z.push_back(i1 + 1), reverse(a, a + i1 + 1);
    if (i2 < i1) i2 = i1 - i2;
    i1 = 0;
    z.push_back(i2), reverse(a, a + i2);
    i1 = i2 - 1;
    z.push_back(i2 + 2), reverse(a, a + i2 + 2);
    z.push_back(3), reverse(a, a + 3);
    z.push_back(n - o + 1), reverse(a, a + n - o + 1);
    o += 2;
  }
  cout << z.size() + 1 << '\n';
  for (int x : z) cout << x << ' ';
  cout << n << '\n';
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  int T = 1;
  cin >> T;
  for (int K = 0; K < T; K++) Darling(K + 1);
}

### 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;
const long long mod = 1000000007, inf = 2e18;
const int N = 200001;
const double ep = 1e-10;
void Darling(int kase) {
  int n;
  cin >> n;
  int a[n];
  for (int i = 0; (1 < 0 && i > n) || (1 > 0 && i < n); i += 1) cin >> a[i];
  for (int i = 0; (1 < 0 && i > n) || (1 > 0 && i < n); i += 1)
    if (a[i] % 2 == i % 2) {
      cout << -1 << '\n';
      return;
    }
  vector<int> z;
  int o = 1;
  while (o < n) {
    int i1 = 0, i2 = 1;
    for (int i = 0; (1 < 0 && i > n) || (1 > 0 && i < n); i += 1) {
      if (a[i] == o) i1 = i;
      if (a[i] == o + 1) i2 = i;
    }
    z.push_back(i1 + 1), reverse(a, a + i1 + 1);
    if (i2 < i1) i2 = i1 - i2;
    i1 = 0;
    z.push_back(i2), reverse(a, a + i2);
    i1 = i2 - 1;
    z.push_back(i2 + 2), reverse(a, a + i2 + 2);
    z.push_back(3), reverse(a, a + 3);
    z.push_back(n - o + 1), reverse(a, a + n - o + 1);
    o += 2;
  }
  cout << z.size() + 1 << '\n';
  for (int x : z) cout << x << ' ';
  cout << n << '\n';
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  int T = 1;
  cin >> T;
  for (int K = 0; K < T; K++) Darling(K + 1);
}

```