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;
  const int MAXN = 2100;
  int n;
  int p[MAXN];
  vector<int> ans;
  void rev(int x) {
    if (x == 1) return;
    if (ans.empty() == false && ans.back() == x) {
      ans.pop_back();
      return;
    }
    ans.push_back(x);
    reverse(p + 1, p + x + 1);
  }
  void rev(int x, int &ind, int &ind1) {
    rev(x);
    if (ind <= x) ind = x - ind + 1;
    if (ind1 <= x) ind1 = x - ind1 + 1;
  }
  void solveTestcase() {
    ans.clear();
    int n;
    cin >> n;
    for (int i = 1; i <= n; i++) cin >> p[i];
    for (int i = 1; i <= n; i++) {
      if (p[i] % 2 != i % 2) {
        cout << "-1" << '\n';
        return;
      }
    }
    for (int num = n; num > 1; num -= 2) {
      if (p[num] == num && p[num - 1] == num - 1) continue;
      if (p[1] == num && p[2] == num - 1) {
        rev(num);
        continue;
      }
      int ind = -1, ind1 = -1;
      for (int i = 1; i <= num; i++) {
        if (p[i] == num) {
          ind = i;
        }
        if (p[i] == num - 1) {
          ind1 = i;
        }
      }
      if (p[ind - 1] == num - 1) {
        rev(ind);
        rev(num);
        continue;
      }
      if (p[ind + 1] == num - 1) {
        rev(num);
        ind = num - ind + 1;
        rev(ind);
        rev(num);
        continue;
      }
      if (ind < ind1) {
        rev(ind, ind, ind1);
        rev(ind1 - 1, ind, ind1);
        rev(num, ind, ind1);
        rev(ind);
        rev(num);
        continue;
      }
      if (ind > ind1) {
        rev(ind, ind, ind1);
        rev(ind1 - 1, ind, ind1);
        rev(num, ind, ind1);
        rev(ind, ind, ind1);
        rev(num, ind, ind1);
        continue;
      }
      cout << "-1" << '\n';
      return;
    }
    if (ans.size() > n * 2.5) {
      cout << "-1" << '\n';
      return;
    }
    cout << ans.size() << '\n';
    for (int x : ans) cout << x << " ";
    cout << '\n';
  }
  int main() {
    int T;
    cin >> T;
    while (T--) solveTestcase();
  }
  