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;
  template <class T>
  using V = vector<T>;
  template <class T>
  using VV = V<V<T>>;
  template <typename T>
  void view(T e) {
    std::cerr << e << std::endl;
  }
  template <typename T>
  void view(const std::vector<T> &v) {
    for (const auto &e : v) {
      std::cerr << e << " ";
    }
    std::cerr << std::endl;
  }
  template <typename T>
  void view(const std::vector<std::vector<T>> &vv) {
    cerr << endl;
    int cnt = 0;
    for (const auto &v : vv) {
      cerr << cnt << "th : ";
      view(v);
      cnt++;
    }
    cerr << endl;
  }
  std::ostream &operator<<(std::ostream &dest, __int128_t value) {
    std::ostream::sentry s(dest);
    if (s) {
      __uint128_t tmp = value < 0 ? -value : value;
      char buffer[128];
      char *d = std::end(buffer);
      do {
        --d;
        *d = "0123456789"[tmp % 10];
        tmp /= 10;
      } while (tmp != 0);
      if (value < 0) {
        --d;
        *d = '-';
      }
      int len = std::end(buffer) - d;
      if (dest.rdbuf()->sputn(d, len) != len) {
        dest.setstate(std::ios_base::badbit);
      }
    }
    return dest;
  }
  ll power(ll a, ll p) {
    ll ret = 1;
    while (p) {
      if (p & 1) {
        ret = ret * a;
      }
      a = a * a;
      p >>= 1;
    }
    return ret;
  }
  ll modpow(ll a, ll p, ll mod) {
    ll ret = 1;
    while (p) {
      if (p & 1) {
        ret = ret * a % mod;
      }
      a = a * a % mod;
      p >>= 1;
    }
    return ret;
  }
  template <class T, class K>
  bool chmax(T &a, const K b) {
    if (a < b) {
      a = b;
      return 1;
    }
    return 0;
  }
  template <class T, class K>
  bool chmin(T &a, const K b) {
    if (b < a) {
      a = b;
      return 1;
    }
    return 0;
  }
  int dx[] = {1, 0, -1, 0};
  int dy[] = {0, 1, 0, -1};
  const int inf = 1001001001;
  const ll INF = 1001001001001001001ll;
  int main() {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    int t;
    cin >> t;
    while (t--) {
      int n;
      cin >> n;
      V<int> a(n);
      for (ll i = 0; i < n; ++i) cin >> a[i];
      bool ok = true;
      for (ll i = 0; i < n; ++i) {
        if (a[i] % 2 == 1 && i % 2 == 1 || a[i] % 2 == 0 && i % 2 == 0) {
          cout << -1 << endl;
          ok = false;
          break;
        }
      }
      if (!ok) continue;
      V<int> ans;
      for (int i = 1; i < n; i += 2) {
        int p, q;
        for (ll j = 0; j < n; ++j) {
          if (a[j] == i) p = j;
        }
        if (p != 0) ans.push_back(p + 1);
        reverse(a.begin(), a.begin() + p + 1);
        for (ll j = 0; j < n; ++j) {
          if (a[j] == i) p = j;
          if (a[j] == i + 1) q = j;
        }
        ans.push_back(q);
        reverse(a.begin(), a.begin() + q);
        ans.push_back(q + 2);
        reverse(a.begin(), a.begin() + q + 2);
        ans.push_back(3);
        reverse(a.begin(), a.begin() + 3);
        ans.push_back(n - i + 1);
        reverse(a.begin(), a.end() - i + 1);
      }
      ans.push_back(n);
      cout << ((int)(ans).size()) << endl;
      for (ll i = 0; i < ((int)(ans).size()); ++i) cout << ans[i] << " ";
      cout << endl;
    }
  }
  