#include <bits/stdc++.h>
using namespace std;
const int N = 2027;
template <typename T>
void read(T &x) {
  bool neg = false;
  unsigned char c = getchar();
  for (; (c ^ 48) > 9; c = getchar())
    if (c == '-') neg = true;
  for (x = 0; (c ^ 48) < 10; c = getchar()) x = (x << 3) + (x << 1) + (c ^ 48);
  if (neg) x = -x;
}
int t, n, a[N], m, ans[5 * N];
bool check() {
  for (int i = 0, i_end = n; i < i_end; ++i)
    if ((i & 1) != (a[i] & 1)) return false;
  return true;
}
void op(int len) {
  for (int i = 0, i_end = len / 2; i < i_end; ++i)
    a[i] ^= a[len - 1 - i] ^= a[i] ^= a[len - 1 - i];
  ans[m++] = len;
}
void solve(int n) {
  int apos, bpos;
  if (n == 1) return;
  for (int i = 0, i_end = n; i < i_end; ++i)
    if (a[i] == n - 1) bpos = i;
  op(++bpos);
  for (int i = 0, i_end = n; i < i_end; ++i)
    if (a[i] == n - 2) apos = i;
  op(apos);
  op(apos + 2);
  op(3);
  op(n);
  solve(n - 2);
}
int main() {
  read(t);
  while (t--) {
    read(n);
    for (int i = 0, i_end = n; i < i_end; ++i) read(a[i]), --a[i];
    if (check()) {
      m = 0;
      solve(n);
      printf("%d\n", m);
      for (int i = 0, i_end = m; i < i_end; ++i) printf("%d ", ans[i]);
      puts("");
    } else
      puts("-1");
  }
  return 0;
}

### Prompt

Generate 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 N = 2027;
template <typename T>
void read(T &x) {
  bool neg = false;
  unsigned char c = getchar();
  for (; (c ^ 48) > 9; c = getchar())
    if (c == '-') neg = true;
  for (x = 0; (c ^ 48) < 10; c = getchar()) x = (x << 3) + (x << 1) + (c ^ 48);
  if (neg) x = -x;
}
int t, n, a[N], m, ans[5 * N];
bool check() {
  for (int i = 0, i_end = n; i < i_end; ++i)
    if ((i & 1) != (a[i] & 1)) return false;
  return true;
}
void op(int len) {
  for (int i = 0, i_end = len / 2; i < i_end; ++i)
    a[i] ^= a[len - 1 - i] ^= a[i] ^= a[len - 1 - i];
  ans[m++] = len;
}
void solve(int n) {
  int apos, bpos;
  if (n == 1) return;
  for (int i = 0, i_end = n; i < i_end; ++i)
    if (a[i] == n - 1) bpos = i;
  op(++bpos);
  for (int i = 0, i_end = n; i < i_end; ++i)
    if (a[i] == n - 2) apos = i;
  op(apos);
  op(apos + 2);
  op(3);
  op(n);
  solve(n - 2);
}
int main() {
  read(t);
  while (t--) {
    read(n);
    for (int i = 0, i_end = n; i < i_end; ++i) read(a[i]), --a[i];
    if (check()) {
      m = 0;
      solve(n);
      printf("%d\n", m);
      for (int i = 0, i_end = m; i < i_end; ++i) printf("%d ", ans[i]);
      puts("");
    } else
      puts("-1");
  }
  return 0;
}

```