#include <bits/stdc++.h>
using namespace std;
template <typename T>
using minpq = priority_queue<T, vector<T>, greater<T>>;
const int INF = 1005000000;
const long long MOD = 1000000007LL;
int ni() {
  int x;
  cin >> x;
  return x;
}
long long nl() {
  long long x;
  cin >> x;
  return x;
}
double nd() {
  double x;
  cin >> x;
  return x;
}
string next() {
  string x;
  cin >> x;
  return x;
}
int search(vector<int> &nums, int i) {
  return find((nums).begin(), (nums).end(), i) - nums.begin();
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int Q = ni();
  for (int q = 0; q < (Q); q++) {
    int N = ni();
    vector<int> nums(N);
    for (int i = 0; i < (N); i++) nums[i] = ni();
    bool good = true;
    for (int i = 1; i <= N; i++) {
      int n = nums[i - 1];
      if ((n % 2) != (i % 2)) {
        good = false;
        break;
      }
    }
    if (!good) {
      cout << "-1\n";
      continue;
    }
    vector<int> ops;
    for (int i = N - 1; i >= 2; i -= 2) {
      int even = search(nums, i);
      int odd = search(nums, i + 1);
      ops.push_back(odd);
      reverse(nums.begin(), nums.begin() + odd + 1);
      even = search(nums, i);
      odd = search(nums, i + 1);
      ops.push_back(even - 1);
      reverse(nums.begin(), nums.begin() + even);
      even = search(nums, i);
      odd = search(nums, i + 1);
      ops.push_back(even + 1);
      reverse(nums.begin(), nums.begin() + even + 2);
      ops.push_back(2);
      reverse(nums.begin(), nums.begin() + 3);
      ops.push_back(i);
      reverse(nums.begin(), nums.begin() + i + 1);
    }
    cout << ((int)(ops).size()) << '\n';
    for (int n : ops) cout << n + 1 << ' ';
    cout << '\n';
  }
}

### 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;
template <typename T>
using minpq = priority_queue<T, vector<T>, greater<T>>;
const int INF = 1005000000;
const long long MOD = 1000000007LL;
int ni() {
  int x;
  cin >> x;
  return x;
}
long long nl() {
  long long x;
  cin >> x;
  return x;
}
double nd() {
  double x;
  cin >> x;
  return x;
}
string next() {
  string x;
  cin >> x;
  return x;
}
int search(vector<int> &nums, int i) {
  return find((nums).begin(), (nums).end(), i) - nums.begin();
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int Q = ni();
  for (int q = 0; q < (Q); q++) {
    int N = ni();
    vector<int> nums(N);
    for (int i = 0; i < (N); i++) nums[i] = ni();
    bool good = true;
    for (int i = 1; i <= N; i++) {
      int n = nums[i - 1];
      if ((n % 2) != (i % 2)) {
        good = false;
        break;
      }
    }
    if (!good) {
      cout << "-1\n";
      continue;
    }
    vector<int> ops;
    for (int i = N - 1; i >= 2; i -= 2) {
      int even = search(nums, i);
      int odd = search(nums, i + 1);
      ops.push_back(odd);
      reverse(nums.begin(), nums.begin() + odd + 1);
      even = search(nums, i);
      odd = search(nums, i + 1);
      ops.push_back(even - 1);
      reverse(nums.begin(), nums.begin() + even);
      even = search(nums, i);
      odd = search(nums, i + 1);
      ops.push_back(even + 1);
      reverse(nums.begin(), nums.begin() + even + 2);
      ops.push_back(2);
      reverse(nums.begin(), nums.begin() + 3);
      ops.push_back(i);
      reverse(nums.begin(), nums.begin() + i + 1);
    }
    cout << ((int)(ops).size()) << '\n';
    for (int n : ops) cout << n + 1 << ' ';
    cout << '\n';
  }
}

```