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;
  int a[2022];
  inline void Solve() {
    int n;
    cin >> n;
    for (register int i = 1; i <= n; i++) {
      cin >> a[i];
    }
    vector<int> A;
    for (int i = n; i != 1; i -= 2) {
      int p1, p2;
      for (register int j = 1; j <= i; j++) {
        if (a[j] == i) {
          p1 = j;
        } else if (a[j] == i - 1) {
          p2 = j;
        }
      }
      if ((p1 & 1) == 0 || (p2 & 1) == 1) {
        cout << -1 << endl;
        return;
      }
      A.push_back(p1);
      reverse(a + 1, a + 1 + p1);
      if (p2 < p1) {
        p2 = p1 - p2 + 1;
      }
      A.push_back(p2 - 1);
      reverse(a + 1, a + p2);
      A.push_back(p2 + 1);
      reverse(a + 1, a + p2 + 2);
      A.push_back(3);
      reverse(a + 1, a + 4);
      A.push_back(i);
      reverse(a + 1, a + i + 1);
    }
    cout << A.size() << endl;
    for (vector<int>::iterator T = A.begin(); T != A.end(); T++) {
      cout << *T << ' ';
    }
    cout << endl;
  }
  int main() {
    int t;
    cin >> t;
    for (register int i = 0; i != t; i++) {
      Solve();
    }
    return 0;
  }
  