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;
  #pragma GCC target("avx2")
  #pragma GCC optimization("O3")
  #pragma GCC optimization("unroll-loops")
  const int tam = 100010;
  const int MOD = 1000000007;
  const int MOD1 = 998244353;
  const double DINF = 1e100;
  const double EPS = 1e-9;
  const double PI = acos(-1);
  void solve() {
    int n;
    cin >> n;
    vector<int> ar(n);
    bool bo = true;
    for (int i = 0; i < n; i++) {
      cin >> ar[i];
      bo &= ar[i] & 1 ^ i;
    }
    if (!bo) {
      cout << -1 << '\n';
      return;
    }
    int a, b;
    vector<int> res;
    for (int i = n; i > 1; i -= 2) {
      for (int j = 0; j < i; j++) {
        if (ar[j] == i) a = j + 1;
        if (ar[j] == i - 1) b = j + 1;
      }
      if (a == i && b == i - 1) {
        continue;
      }
      if (a != 1) {
        res.push_back(a);
        reverse(ar.begin(), ar.begin() + a);
        for (int j = 0; j < i; j++)
          if (ar[j] == i - 1) b = j + 1;
      }
      if (b != 2) {
        res.push_back(b - 1);
        reverse(ar.begin(), ar.begin() + b - 1);
        res.push_back(b + 1);
        reverse(ar.begin(), ar.begin() + b + 1);
        res.push_back(3);
        reverse(ar.begin(), ar.begin() + 3);
      }
      res.push_back(i);
      reverse(ar.begin(), ar.begin() + i);
    }
    cout << res.size() << '\n';
    for (int x : res) cout << x << ' ';
    if (res.size()) cout << '\n';
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int t;
    cin >> t;
    while (t--) {
      solve();
    }
    return 0;
  }
  