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 "Ofast"
  using namespace std;
  int t, n, a[2100];
  vector<int> ans;
  void rev(int x) {
    reverse(a + 1, a + x + 1);
    ans.push_back(x);
  }
  bool check() {
    int ret = 0;
    vector<int> tmp;
    for (int i = 0; i < ans.size(); i++) {
      if (i != ans.size() - 1 and ans[i] == ans[i + 1]) {
        i++;
        ret = 1;
      } else {
        tmp.push_back(ans[i]);
      }
    }
    swap(ans, tmp);
    return ret;
  }
  int main() {
    cin >> t;
    while (t--) {
      bool imp = 0;
      cin >> n;
      ans.clear();
      for (int i = 1; i <= n; i++) {
        cin >> a[i];
        if (a[i] % 2 != i % 2) imp = 1;
      }
      if (imp) {
        cout << "-1\n";
        continue;
      }
      for (int i = n; i >= 3; i -= 2) {
        int pos = 0, pos2 = 0;
        for (int j = 1; j <= n; j++) {
          if (a[j] == i) pos = j;
        }
        for (int j = 1; j <= n; j++) {
          if (a[j] == i - 1) pos2 = j;
        }
        int pos3 = 0;
        for (int j = 1; j <= n; j++) {
          if (a[j] == i) pos3 = j;
        }
        rev(pos3);
        int pos4 = 0;
        for (int j = 1; j <= n; j++) {
          if (a[j] == i - 1) pos4 = j;
        }
        rev(pos4 - 1);
        rev(pos4 + 1);
        rev(3);
        rev(i);
      }
      for (int i = 1; i <= n; i++) {
        if (a[i] != i) {
          imp = 1;
          break;
        }
      }
      if (imp)
        cout << "-1\n";
      else {
        while (check()) {
        }
        if (ans.size() > (5 * n) / 2)
          cout << "-1\n";
        else {
          cout << ans.size() << '\n';
          for (int i = 0; i < ans.size(); i++)
            cout << ans[i] << " \n"[i == ans.size() - 1];
        }
      }
    }
  }
  