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")
  #pragma GCC optimize("unroll-loops")
  using namespace std;
  using ll = long long;
  template <class t, class u>
  bool chmax(t& first, u second) {
    if (first < second) {
      first = second;
      return true;
    } else
      return false;
  }
  template <class t, class u>
  bool chmin(t& first, u second) {
    if (second < first) {
      first = second;
      return true;
    } else
      return false;
  }
  template <class t>
  using vc = vector<t>;
  template <class t>
  using vvc = vc<vc<t>>;
  using pi = pair<ll, ll>;
  using vi = vc<ll>;
  template <class t, class u>
  ostream& operator<<(ostream& os, const pair<t, u>& p) {
    return os << "{" << p.first << "," << p.second << "}";
  }
  template <class t>
  ostream& operator<<(ostream& os, const vc<t>& v) {
    os << "{";
    for (auto e : v) os << e << ",";
    return os << "}";
  }
  using uint = unsigned;
  using ull = unsigned long long;
  template <class t, size_t n>
  ostream& operator<<(ostream& os, const array<t, n>& first) {
    return os << vc<t>(first.begin(), first.end());
  }
  template <ll i, class T>
  void print_tuple(ostream&, const T&) {}
  template <ll i, class T, class H, class... Args>
  void print_tuple(ostream& os, const T& t) {
    if (i) os << ",";
    os << get<i>(t);
    print_tuple<i + 1, T, Args...>(os, t);
  }
  template <class... Args>
  ostream& operator<<(ostream& os, const tuple<Args...>& t) {
    os << "{";
    print_tuple<0, tuple<Args...>, Args...>(os, t);
    return os << "}";
  }
  template <class t>
  void print(t x, ll suc = 1) {
    cout << x;
    if (suc == 1) cout << "\n";
    if (suc == 2) cout << " ";
  }
  ll read() {
    ll i;
    cin >> i;
    return i;
  }
  vi readvi(ll n, ll off = 0) {
    vi v(n);
    for (ll i = ll(0); i < ll(n); i++) v[i] = read() + off;
    return v;
  }
  pi readpi(ll off = 0) {
    ll first, second;
    cin >> first >> second;
    return pi(first + off, second + off);
  }
  template <class t, class u>
  void print(const pair<t, u>& p, ll suc = 1) {
    print(p.first, 2);
    print(p.second, suc);
  }
  template <class T>
  void print(const vector<T>& v, ll suc = 1) {
    for (ll i = ll(0); i < ll(v.size()); i++)
      print(v[i], i == ll(v.size()) - 1 ? suc : 2);
  }
  template <class T>
  void print_offset(const vector<T>& v, ll off, ll suc = 1) {
    for (ll i = ll(0); i < ll(v.size()); i++)
      print(v[i] + off, i == ll(v.size()) - 1 ? suc : 2);
  }
  template <class T, size_t N>
  void print(const array<T, N>& v, ll suc = 1) {
    for (ll i = ll(0); i < ll(N); i++) print(v[i], i == ll(N) - 1 ? suc : 2);
  }
  string readString() {
    string s;
    cin >> s;
    return s;
  }
  template <class T>
  T sq(const T& t) {
    return t * t;
  }
  void yes(bool ex = true) {
    cout << "Yes"
         << "\n";
    if (ex) exit(0);
  }
  void no(bool ex = true) {
    cout << "No"
         << "\n";
    if (ex) exit(0);
  }
  void possible(bool ex = true) {
    cout << "Possible"
         << "\n";
    if (ex) exit(0);
  }
  void impossible(bool ex = true) {
    cout << "Impossible"
         << "\n";
    if (ex) exit(0);
  }
  constexpr ll ten(ll n) { return n == 0 ? 1 : ten(n - 1) * 10; }
  const ll infLL = LLONG_MAX / 3;
  const ll inf = infLL;
  ll topbit(signed t) { return t == 0 ? -1 : 31 - __builtin_clz(t); }
  ll topbit(ll t) { return t == 0 ? -1 : 63 - __builtin_clzll(t); }
  ll botbit(signed first) { return first == 0 ? 32 : __builtin_ctz(first); }
  ll botbit(ll first) { return first == 0 ? 64 : __builtin_ctzll(first); }
  ll popcount(signed t) { return __builtin_popcount(t); }
  ll popcount(ll t) { return __builtin_popcountll(t); }
  bool ispow2(ll i) { return i && (i & -i) == i; }
  ll mask(ll i) { return (ll(1) << i) - 1; }
  bool inc(ll first, ll second, ll c) { return first <= second && second <= c; }
  template <class t>
  void mkuni(vc<t>& v) {
    sort(v.begin(), v.end());
    v.erase(unique(v.begin(), v.end()), v.end());
  }
  ll rand_int(ll l, ll r) {
    static mt19937_64 gen(chrono::steady_clock::now().time_since_epoch().count());
    return uniform_int_distribution<ll>(l, r)(gen);
  }
  template <class t>
  void myshuffle(vc<t>& first) {
    for (ll i = ll(0); i < ll(ll(first.size())); i++)
      swap(first[i], first[rand_int(0, i)]);
  }
  template <class t>
  ll lwb(const vc<t>& v, const t& first) {
    return lower_bound(v.begin(), v.end(), first) - v.begin();
  }
  vvc<ll> readGraph(ll n, ll m) {
    vvc<ll> g(n);
    for (ll i = ll(0); i < ll(m); i++) {
      ll first, second;
      cin >> first >> second;
      first--;
      second--;
      g[first].push_back(second);
      g[second].push_back(first);
    }
    return g;
  }
  vvc<ll> readTree(ll n) { return readGraph(n, n - 1); }
  void slv() {
    ll n;
    cin >> n;
    vi first = readvi(n, -1);
    for (ll i = ll(0); i < ll(n); i++) {
      if (first[i] % 2 != i % 2) {
        print(-1);
        return;
      }
    }
    vi ans;
    auto rev = [&](ll len) {
      assert(inc(1, len, n));
      assert(len % 2);
      if (len == 1) return;
      reverse(first.begin(), first.begin() + len);
      ans.push_back(len);
    };
    ll m = n;
    while (m > 1) {
      {
        ll f = find(first.begin(), first.begin() + m, m - 1) - first.begin();
        assert(f % 2 == 0);
        rev(f + 1);
      }
      {
        ll f = find(first.begin(), first.begin() + m, m - 2) - first.begin();
        assert(f % 2 == 1);
        rev(f);
      }
      rev(m);
      {
        ll f = find(first.begin(), first.begin() + m, m - 1) - first.begin();
        assert(f % 2 == 0);
        rev(f + 1);
      }
      rev(m);
      m -= 2;
    }
    assert(is_sorted(first.begin(), first.end()));
    print(ll(ans.size()));
    print(ans);
  }
  signed main() {
    cin.tie(0);
    ios::sync_with_stdio(0);
    cout << fixed << setprecision(20);
    ll t;
    cin >> t;
    for (ll _ = ll(0); _ < ll(t); _++) slv();
  }
  