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")
  using namespace std;
  long long max(long long a, long long b) {
    if (a > b)
      return a;
    else
      return b;
  }
  long long min(long long a, long long b) {
    if (a > b)
      return b;
    else
      return a;
  }
  long long gcd(long long a, long long b) { return (b == 0) ? a : gcd(b, a % b); }
  inline long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
  inline long long normalize(long long x, long long mod) {
    x %= mod;
    if (x < 0) x += mod;
    return x;
  }
  long long dx[4] = {0, 1, 0, -1};
  long long dy[4] = {1, 0, -1, 0};
  string abc = "abcdefghijklmnopqrstuvwxyz";
  const double PI = acosl(-1);
  const long long mod = 1000 * 1000 * 1000 + 7;
  const long long inf = 1e18 + 5;
  inline long long add(long long a, long long b, long long p = mod) {
    long long c = a + b;
    if (c >= p) c -= p;
    return c;
  }
  inline long long sub(long long a, long long b, long long p = mod) {
    long long c = a - b;
    if (c < 0) c += p;
    return c;
  }
  inline long long mul(long long a, long long b, long long p = mod) {
    return (a * 1LL * b) % p;
  }
  long long floor1(long long n, long long k) {
    if (n % k == 0 || n >= 0) return n / k;
    return (n / k) - 1;
  }
  long long ceil1(long long n, long long k) { return floor1(n + k - 1, k); }
  long long powm(long long a, long long b) {
    long long res = 1;
    while (b) {
      if (b & 1) res = (res * a) % mod;
      a = (a * a) % mod;
      b >>= 1;
    }
    return res;
  }
  const long long N = 2021 + 5;
  long long n;
  long long a[N], f[N];
  vector<long long> ans;
  void op(long long l, long long r) {
    ans.push_back(r);
    while (l <= r) {
      swap(f[a[l]], f[a[r]]);
      swap(a[l], a[r]);
      ++l;
      --r;
    }
  }
  void solve() {
    cin >> n;
    for (long long i = (1); i < (n + 1); ++i) cin >> a[i];
    for (long long i = (1); i < (n + 1); ++i) f[a[i]] = i;
    ans.clear();
    for (long long i = n; i > 1; i -= 2) {
      long long p = f[i];
      if (p == i and i - 1 == f[i - 1]) continue;
      if (p % 2 == 0 or f[i - 1] % 2 == 1) {
        cout << -1 << "\n";
        return;
      }
      op(1, f[i]);
      op(1, f[i - 1] - 1);
      op(1, f[i - 1] + 1);
      op(1, 3);
      op(1, i);
    }
    cout << ((long long)ans.size()) << "\n";
    for (long long i : ans) cout << i << " ";
    cout << "\n";
  }
  signed main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    cout << fixed << setprecision(10);
    long long t = 1;
    cin >> t;
    for (long long i = (1); i < (t + 1); ++i) {
      solve();
    }
    return 0;
  }
  