#include <bits/stdc++.h>
using namespace std;
const long long M = 1e9 + 7;
long long pmp(long long x, long long n) {
  long long a = 1;
  while (n > 0) {
    if (n & 1) a = (a * x) % M;
    x = (x * x) % M;
    n = n >> 1;
  }
  return a;
}
void solve() {
  long long n;
  cin >> n;
  vector<long long> a(n);
  vector<long long> inv(n);
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    --a[i];
    inv[a[i]] = i;
  }
  for (long long i = 0; i < n; i++) {
    if (i % 2 != a[i] % 2) {
      cout << "-1\n";
      return;
    }
  }
  vector<long long> op;
  long long nxt = n - 1;
  while (nxt) {
    long long last = inv[nxt];
    long long prev = inv[nxt - 1];
    if (last == nxt && prev == nxt - 1) {
      nxt -= 2;
      continue;
    }
    if (last > prev || last) {
      op.push_back(last + 1);
      reverse(a.begin(), a.begin() + last + 1);
      prev = (last > prev ? last - prev : prev);
      last = 0;
    }
    if (prev - 1) {
      op.push_back(prev);
      reverse(a.begin(), a.begin() + prev);
      last = prev - 1 - last;
    }
    if (last < prev) {
      op.push_back(prev + 2);
      reverse(a.begin(), a.begin() + prev + 2);
    }
    op.push_back(3);
    reverse(a.begin(), a.begin() + 3);
    op.push_back(nxt + 1);
    reverse(a.begin(), a.begin() + nxt + 1);
    nxt -= 2;
    for (long long i = 0; i < nxt + 1; i++) inv[a[i]] = i;
  }
  cout << op.size() << "\n";
  for (auto x : op) cout << x << " ";
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t;
  t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  cerr << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC
       << "ms\n";
  return 0;
}

### Prompt

Create a solution in CPP for 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 long long M = 1e9 + 7;
long long pmp(long long x, long long n) {
  long long a = 1;
  while (n > 0) {
    if (n & 1) a = (a * x) % M;
    x = (x * x) % M;
    n = n >> 1;
  }
  return a;
}
void solve() {
  long long n;
  cin >> n;
  vector<long long> a(n);
  vector<long long> inv(n);
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    --a[i];
    inv[a[i]] = i;
  }
  for (long long i = 0; i < n; i++) {
    if (i % 2 != a[i] % 2) {
      cout << "-1\n";
      return;
    }
  }
  vector<long long> op;
  long long nxt = n - 1;
  while (nxt) {
    long long last = inv[nxt];
    long long prev = inv[nxt - 1];
    if (last == nxt && prev == nxt - 1) {
      nxt -= 2;
      continue;
    }
    if (last > prev || last) {
      op.push_back(last + 1);
      reverse(a.begin(), a.begin() + last + 1);
      prev = (last > prev ? last - prev : prev);
      last = 0;
    }
    if (prev - 1) {
      op.push_back(prev);
      reverse(a.begin(), a.begin() + prev);
      last = prev - 1 - last;
    }
    if (last < prev) {
      op.push_back(prev + 2);
      reverse(a.begin(), a.begin() + prev + 2);
    }
    op.push_back(3);
    reverse(a.begin(), a.begin() + 3);
    op.push_back(nxt + 1);
    reverse(a.begin(), a.begin() + nxt + 1);
    nxt -= 2;
    for (long long i = 0; i < nxt + 1; i++) inv[a[i]] = i;
  }
  cout << op.size() << "\n";
  for (auto x : op) cout << x << " ";
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t;
  t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  cerr << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC
       << "ms\n";
  return 0;
}

```