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>
  #pragma GCC optimize("O3")
  #pragma GCC target("sse4")
  #pragma GCC optimize("unroll-loops")
  using namespace std;
  template <class T>
  inline void amin(T &x, const T &y) {
    if (y < x) x = y;
  }
  template <class T>
  inline void amax(T &x, const T &y) {
    if (x < y) x = y;
  }
  int N;
  int A[2222];
  vector<int> ans;
  void ADD(int x) {
    assert(x % 2 == 1);
    ans.push_back(x);
    reverse(A, A + x);
  }
  void MAIN() {
    scanf("%d", &N);
    for (int i = 0, i_len = (N); i < i_len; ++i) scanf("%d", A + i);
    bool yes = true;
    for (int i = 0, i_len = (N); i < i_len; ++i)
      if (A[i] % 2 == i % 2) yes = false;
    if (yes) {
      ans.clear();
      int len = N;
      while (len >= 3) {
        if (A[len - 1] == len && A[len - 2] == len - 1) {
          len -= 2;
        } else if (A[0] == len && A[1] == len - 1) {
          ADD(len);
        } else {
          int a = -1, b = -1;
          for (int i = 0, i_len = (len); i < i_len; ++i) {
            if (A[i] == len) a = i;
            if (A[i] == len - 1) b = i;
          }
          if (b + 1 == a) {
            ADD(a + 1);
          } else if (a + 1 == b) {
            ADD(b + 2);
          } else if (a == 0) {
            ADD(b);
          } else {
            ADD(a + 1);
          }
        }
      }
      printf("%d\n", (int)ans.size());
      do {
        const auto end_rp = (ans.end());
        auto it_rp = (ans.begin());
        for (bool sp_rp = 0; it_rp != end_rp; ++it_rp) {
          if (sp_rp)
            putchar(' ');
          else
            sp_rp = true;
          printf("%d", *it_rp);
        }
        putchar('\n');
      } while (0);
    } else {
      puts("-1");
    }
  }
  int main() {
    int TC = 1;
    scanf("%d", &TC);
    for (int tc = 0, tc_len = (TC); tc < tc_len; ++tc) MAIN();
    return 0;
  }
  