#include <bits/stdc++.h>
using namespace std;
int t, n, i, nr, a[2025], poz[2025], b[10004];
bool ok;
void rev(int x) {
  int i;
  if (x == 1) return;
  b[++nr] = x;
  for (i = 1; i <= (x / 2); i++) {
    swap(a[i], a[x - i + 1]);
    swap(poz[a[i]], poz[a[x - i + 1]]);
  }
}
int main() {
  scanf("%d", &t);
  while (t > 0) {
    scanf("%d", &n);
    ok = true;
    for (i = 1; i <= n; i++) {
      scanf("%d", &a[i]);
      if (((a[i] - i) % 2) != 0) ok = false;
      poz[a[i]] = i;
    }
    nr = 0;
    if (ok == true) {
      for (i = 1; i < n; i += 2) {
        rev(poz[i]);
        rev(poz[i + 1] - 1);
        rev(poz[i + 1] + 1);
        rev(3);
        rev(n - i + 1);
      }
      rev(n);
      printf("%d\n", nr);
      for (i = 1; i <= nr; i++) printf("%d ", b[i]);
      printf("\n");
    } else
      printf("%d\n", -1);
    t--;
  }
  return 0;
}

### 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>
using namespace std;
int t, n, i, nr, a[2025], poz[2025], b[10004];
bool ok;
void rev(int x) {
  int i;
  if (x == 1) return;
  b[++nr] = x;
  for (i = 1; i <= (x / 2); i++) {
    swap(a[i], a[x - i + 1]);
    swap(poz[a[i]], poz[a[x - i + 1]]);
  }
}
int main() {
  scanf("%d", &t);
  while (t > 0) {
    scanf("%d", &n);
    ok = true;
    for (i = 1; i <= n; i++) {
      scanf("%d", &a[i]);
      if (((a[i] - i) % 2) != 0) ok = false;
      poz[a[i]] = i;
    }
    nr = 0;
    if (ok == true) {
      for (i = 1; i < n; i += 2) {
        rev(poz[i]);
        rev(poz[i + 1] - 1);
        rev(poz[i + 1] + 1);
        rev(3);
        rev(n - i + 1);
      }
      rev(n);
      printf("%d\n", nr);
      for (i = 1; i <= nr; i++) printf("%d ", b[i]);
      printf("\n");
    } else
      printf("%d\n", -1);
    t--;
  }
  return 0;
}

```