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 pii = pair<int, int>;
  using pll = pair<ll, ll>;
  template <class T>
  using vv = vector<vector<T>>;
  template <class T>
  inline bool MX(T &l, const T &r) {
    return l < r ? l = r, 1 : 0;
  }
  template <class T>
  inline bool MN(T &l, const T &r) {
    return l > r ? l = r, 1 : 0;
  }
  void ouT(ll x, int d = 3) {
    auto re = to_string(x);
    if ((int)re.size() > d) re = x > 0 ? "oo" : "-oo";
    cout << string(d - re.size(), ' ') << re << ",";
  }
  template <class A, class B>
  pair<A, B> operator+(const pair<A, B> &p, const pair<A, B> &q) {
    return {p.first + q.first, p.second + q.second};
  }
  template <class A, class B, class C, class D>
  pair<A, B> &operator+=(pair<A, B> &p, const pair<C, D> &q) {
    p.first += q.first;
    p.second += q.second;
    return p;
  }
  template <class A, class B>
  pair<A, B> operator-(const pair<A, B> &p, const pair<A, B> &q) {
    return {p.first - q.first, p.second - q.second};
  }
  template <class A, class B, class C, class D>
  pair<A, B> &operator-=(pair<A, B> &p, const pair<C, D> &q) {
    p.first -= q.first;
    p.second -= q.second;
    return p;
  }
  template <class A, class B>
  istream &operator>>(istream &is, pair<A, B> &p) {
    is >> p.first >> p.second;
    return is;
  }
  template <class T = ll>
  T read() {
    T re;
    cin >> re;
    return re;
  }
  template <class T = ll>
  T read(const T &dec) {
    T re;
    cin >> re;
    return re - dec;
  }
  template <class T = ll>
  vector<T> readV(const int sz) {
    vector<T> re(sz);
    for (auto &x : re) x = read<T>();
    return re;
  }
  template <class T = ll>
  vector<T> readV(const int sz, const T &dec) {
    vector<T> re(sz);
    for (auto &x : re) x = read<T>(dec);
    return re;
  }
  vv<int> readG(const int &n, const int &m) {
    vv<int> g(n);
    for (int(_) = 0; (_) < (int)(m); ++(_)) {
      const auto a = read<int>(1), b = read<int>(1);
      g[a].push_back(b);
      g[b].push_back(a);
    }
    return g;
  }
  vv<int> readG(const int &n) { return readG(n, n - 1); }
  vv<int> readD(const int &n, const int &m) {
    vv<int> g(n);
    for (int(_) = 0; (_) < (int)(m); ++(_)) {
      const auto a = read<int>(1), b = read<int>(1);
      g[a].push_back(b);
    }
    return g;
  }
  vv<int> readD(const int &n) { return readD(n, n - 1); }
  template <class T>
  vv<pair<int, T>> readG(const int &n, const int &m) {
    vv<pair<int, T>> g(n);
    for (int(_) = 0; (_) < (int)(m); ++(_)) {
      const auto a = read<int>(1), b = read<int>(1);
      const auto c = read<T>();
      g[a].emplace_back(b, c);
      g[b].emplace_back(a, c);
    }
    return g;
  }
  template <class T>
  vv<pair<int, T>> readG(const int &n) {
    return readG<T>(n, n - 1);
  }
  template <class T>
  vv<pair<int, T>> readD(const int &n, const int &m) {
    vv<pair<int, T>> g(n);
    for (int(_) = 0; (_) < (int)(m); ++(_)) {
      const auto a = read<int>(1), b = read<int>(1);
      const auto c = read<T>();
      g[a].emplace_back(b, c);
    }
    return g;
  }
  template <class T>
  vv<pair<int, T>> readD(const int &n) {
    return readD<T>(n, n - 1);
  }
  const ll MOD = 1e9 + 7;
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout << fixed << setprecision(0);
    const auto TTT = read();
    for (int(_) = 0; (_) < (int)(TTT); ++(_)) {
      const auto n = read();
      auto vs = readV(n, 1);
      {
        int ok = 1;
        for (int(i) = 0; (i) < (int)(n); ++(i))
          if ((i + vs[i]) % 2) ok = 0;
        if (!ok) {
          cout << -1 << "\n";
          continue;
        }
      }
      vector<int> re;
      const auto rev = [&](const int &len) {
        re.emplace_back(len);
        reverse(vs.begin(), vs.begin() + len);
        ;
      };
      for (int(v) = (int)(n)-1; (v) >= 0; --(v))
        if (v % 2) {
          const auto a = v;
          const auto b = v + 1;
          ;
          auto ai = find((vs).begin(), (vs).end(), a) - vs.begin();
          auto bi = find((vs).begin(), (vs).end(), b) - vs.begin();
          if (abs(ai - bi) > 1) {
            rev(bi + 1);
            ai = find((vs).begin(), (vs).end(), a) - vs.begin();
            rev(ai);
            bi = ai - 1;
          }
          if (bi + 1 == ai) {
            rev(ai + 2);
            ai = 1;
            bi = 2;
          }
          if (ai != a) {
            rev(bi + 1);
            rev(b + 1);
          };
        }
      int ok = 1;
      for (int(i) = 0; (i) < (int)(n); ++(i))
        if (i != vs[i]) ok = 0;
      if (ok) {
        cout << re.size() << "\n";
        for (const auto &x : re) cout << x << " ";
        cout << "\n";
      } else {
        cout << -1 << "\n";
      }
    }
    return 0;
  }
  