description: 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.
solution: #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';
    }
  }
  