#include <bits/stdc++.h>
using namespace std;
const int N = 2021 + 5, M = 200 + 5;
int pos[N], a[N], b[N];
int n;
vector<int> res;
void update(int x) {
  res.push_back(x);
  for (int i = 1; i <= (x + 1) / 2; i++) {
    swap(a[i], a[x - i + 1]);
  }
  for (int i = 1; i <= n; i++) {
    pos[a[i]] = i;
  }
}
bool check() {
  for (int i = 1; i <= n; i++) {
    if (a[i] != i) return false;
  }
  return true;
}
void solve() {
  memset(pos, 0, sizeof pos);
  memset(a, 0, sizeof a);
  res.clear();
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    pos[a[i]] = i;
    b[i] = a[i];
  }
  for (int i = 1; i <= n; i++) {
    if ((a[i] & 1) != (i & 1)) {
      puts("-1");
      return;
    }
  }
  for (int i = n; i >= 3; i -= 2) {
    int x = i, y = i - 1;
    if (pos[x] > pos[y]) update(pos[x]);
    if (pos[x] + 1 != pos[y] && pos[x] != 1) update(pos[x]);
    if (pos[x] + 1 != pos[y]) update(pos[y] - 1);
    if (pos[x] != 1) {
      update(pos[y] + 1);
      update(pos[x]);
    }
    update(x);
    if (check()) break;
  }
  if (res.size() >= 5 * n / 2) {
    puts("-1");
    return;
  }
  printf("%d\n", res.size());
  for (auto x : res) {
    printf("%d ", x);
  }
  puts("");
}
int main() {
  int t;
  cin >> t;
  while (t--) solve();
  return 0;
}

### Prompt

Please formulate 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 = 2021 + 5, M = 200 + 5;
int pos[N], a[N], b[N];
int n;
vector<int> res;
void update(int x) {
  res.push_back(x);
  for (int i = 1; i <= (x + 1) / 2; i++) {
    swap(a[i], a[x - i + 1]);
  }
  for (int i = 1; i <= n; i++) {
    pos[a[i]] = i;
  }
}
bool check() {
  for (int i = 1; i <= n; i++) {
    if (a[i] != i) return false;
  }
  return true;
}
void solve() {
  memset(pos, 0, sizeof pos);
  memset(a, 0, sizeof a);
  res.clear();
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    pos[a[i]] = i;
    b[i] = a[i];
  }
  for (int i = 1; i <= n; i++) {
    if ((a[i] & 1) != (i & 1)) {
      puts("-1");
      return;
    }
  }
  for (int i = n; i >= 3; i -= 2) {
    int x = i, y = i - 1;
    if (pos[x] > pos[y]) update(pos[x]);
    if (pos[x] + 1 != pos[y] && pos[x] != 1) update(pos[x]);
    if (pos[x] + 1 != pos[y]) update(pos[y] - 1);
    if (pos[x] != 1) {
      update(pos[y] + 1);
      update(pos[x]);
    }
    update(x);
    if (check()) break;
  }
  if (res.size() >= 5 * n / 2) {
    puts("-1");
    return;
  }
  printf("%d\n", res.size());
  for (auto x : res) {
    printf("%d ", x);
  }
  puts("");
}
int main() {
  int t;
  cin >> t;
  while (t--) solve();
  return 0;
}

```