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;
  using ll = long long;
  using matr = vector<vector<ll>>;
  using dbl = long double;
  using ui = unsigned int;
  using ld = long double;
  mt19937 rnd;
  ll random(ll limit) {
    uniform_int_distribution<ll> uniform_dist(0, limit);
    return uniform_dist(rnd);
  }
  ll mod = 998244353;
  ll bpow(ll a, ll p) {
    ll res = 1;
    while (p > 0) {
      if (p % 2) res *= a;
      p /= 2;
      a *= a;
      res %= mod;
      a %= mod;
    }
    return res;
  }
  ll inv(ll a) { return bpow(a, mod - 2); }
  ll gcd(ll a, ll b) {
    if (!b) return a;
    return gcd(b, a % b);
  }
  vector<int> ans;
  void lol(vector<int>& x, int a) {
    for (int i = 0; i < (a - 1) / 2; i++) {
      swap(x[i], x[a - i - 1]);
    }
    ans.push_back(a);
  }
  void mem(vector<int>& x, int& pos1, int& pos2, int cur) {
    pos1 = -1, pos2 = -1;
    for (int i = 0; i < x.size(); i++) {
      if (x[i] == cur) pos1 = i;
      if (x[i] == cur - 1) pos2 = i;
    }
  }
  void solve() {
    int n;
    cin >> n;
    vector<int> x(n);
    bool f = 1;
    for (int i = 0; i < n; i++) {
      cin >> x[i];
      if (x[i] % 2 == i % 2) {
        f = 0;
      }
    }
    if (!f) {
      cout << "-1\n";
      return;
    }
    ans = {};
    for (int cur = n; cur >= 3; cur -= 2) {
      int pos1, pos2;
      mem(x, pos1, pos2, cur);
      lol(x, pos1 + 1);
      mem(x, pos1, pos2, cur);
      lol(x, pos2);
      mem(x, pos1, pos2, cur);
      lol(x, pos2 + 2);
      mem(x, pos1, pos2, cur);
      lol(x, 3);
      lol(x, cur);
    }
    cout << ans.size() << '\n';
    for (auto v : ans) cout << v << ' ';
    cout << '\n';
  }
  int main() {
    int t;
    t = 1;
    cin >> t;
    while (t--) {
      solve();
    }
  }
  