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;
  struct chash {
    static inline uint64_t splitmix64(uint64_t x) {
      x += 0x9e3779b97f4a7c15;
      x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
      x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
      return x ^ (x >> 31);
    }
    inline size_t operator()(uint64_t x) const {
      static const uint64_t FIXED_RANDOM =
          chrono::steady_clock::now().time_since_epoch().count();
      return splitmix64(x + FIXED_RANDOM);
    }
    inline size_t operator()(const pair<uint64_t, uint64_t> &v) const {
      chash int_hasher;
      return int_hasher(v.first) * 31 + int_hasher(v.second);
    }
  };
  const char nl = '\n';
  const long long inf = 1e9 + 10;
  const long long inf2 = 1e18 + 99LL;
  const long double inf3 = 1e17;
  const long long mod = 1e9 + 7, mod2 = 998244353;
  const long double eps = 1e-9;
  const bool local = false;
  const int logn = 21, maxn = 200001, maxm = 1, maxn2 = 3;
  int a[maxn];
  void f(int x) {
    for (int i = 0; i < x / 2; i++) {
      swap(a[i], a[x - i - 1]);
    }
  }
  void solve() {
    int n;
    cin >> n;
    for (int i = 0; i < n; i++) {
      cin >> a[i];
    }
    for (int i = 0; i < n; i++) {
      if (a[i] % 2 != (i + 1) % 2) {
        cout << -1 << nl;
        return;
      }
    }
    vector<int> ans;
    for (int i = n; i > 1; i -= 2) {
      if (a[i - 1] == i && a[i - 2] == i - 1) continue;
      int i1 = 0, i2 = 0;
      for (int j = 0; j < n; j++) {
        if (a[j] == i)
          i1 = j;
        else if (a[j] == i - 1)
          i2 = j;
      }
      if (i1 + 1 > 1) {
        ans.push_back(i1 + 1);
        f(i1 + 1);
        if (i2 < i1) i2 = i1 - i2;
        i1 = 0;
      }
      if (i2 > 1) {
        ans.push_back(i2);
        f(i2);
      }
      i1 = i2 - 1;
      f(i);
      ans.push_back(i);
      i2 = i - i2 - 1;
      f(i2 + 2);
      ans.push_back(i2 + 2);
      f(i);
      ans.push_back(i);
    }
    cout << (int)((ans).size()) << nl;
    for (int x : ans) cout << x << ' ';
    cout << nl;
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    ;
    int T;
    cin >> T;
    while (T--) solve();
  }
  