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;
  using ll = long long;
  using ld = long double;
  const ll MOD = 1e+9 + 7;
  void MAIN() {
    int n;
    cin >> n;
    int arr[n + 1];
    for (int i = 1; i <= n; i++) cin >> arr[i];
    for (int i = 1; i <= n; i++) {
      if (arr[i] % 2 != i % 2) {
        cout << -1 << '\n';
        return;
      }
    }
    auto reverse = [&](int x) {
      for (int i = 1; i <= x / 2; i++) swap(arr[i], arr[x - i + 1]);
    };
    vector<int> ans;
    for (int i = n; i > 1; i--) {
      if (i % 2) {
        int x = find(arr + 1, arr + n + 1, i) - arr;
        ans.push_back(x);
        reverse(x);
        x = find(arr + 1, arr + n + 1, i - 1) - arr;
        reverse(x - 1);
        ans.push_back(x - 1);
        reverse(x + 1);
        ans.push_back(x + 1);
        reverse(3);
        ans.push_back(3);
        reverse(i);
        ans.push_back(i);
      }
    }
    cout << ans.size() << '\n';
    for (int i : ans) cout << i << ' ';
    cout << '\n';
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    cout << fixed;
    cout << setprecision(10);
    int t;
    cin >> t;
    while (t--) MAIN();
  }
  