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 recur_depth = 0;
  const long long sz = 2100;
  long long ara[sz];
  int main() {
    long long t;
    cin >> t;
    while (t--) {
      long long n;
      scanf("%lld", &n);
      bool ok = 1;
      for (long long i = 1; i <= n; ++i) {
        scanf("%lld", &ara[i]);
        if ((ara[i] & 1) != (i & 1)) ok = 0;
      }
      if (!ok) {
        printf("-1\n");
        continue;
      }
      vector<int> ans;
      long long lim = n;
      while (!is_sorted(ara + 1, ara + lim + 1)) {
        for (long long i = 1; i <= lim; ++i) {
          if (ara[i] == lim) {
            ans.push_back(i);
            reverse(ara + 1, ara + i + 1);
            break;
          }
        }
        long long idx;
        for (long long i = 1; i <= lim; ++i) {
          if (ara[i] == lim - 1) {
            ans.push_back(i - 1);
            reverse(ara + 1, ara + i);
            idx = i;
            break;
          }
        }
        idx++;
        ans.push_back(idx);
        reverse(ara + 1, ara + idx + 1);
        ans.push_back(3);
        reverse(ara + 1, ara + 3 + 1);
        ans.push_back(lim);
        reverse(ara + 1, ara + lim + 1);
        lim -= 2;
      }
      printf("%d\n", (int)ans.size());
      for (int& x : ans) printf("%d ", x);
      if (!ans.empty()) printf("\n");
    }
    return 0;
  }
  