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;
  const long long inf = (1ll << 62) - 1;
  const long long mod = 1e9 + 7;
  const long long N = 1e6 + 100;
  const long double pi = 3.14159265358979323846;
  const long double eps = 1e-7;
  long long dx[] = {-1, 0, 1, 0, -1, -1, 1, 1};
  long long dy[] = {0, 1, 0, -1, -1, 1, 1, -1};
  char dir[] = {'U', 'R', 'D', 'L'};
  long long bPow(long long a, long long b) {
    long long res = 1;
    while (b) {
      if (b & 1) {
        res = (res * a) % mod;
      }
      b >>= 1;
      a = (a * a) % mod;
    }
    return res % mod;
  }
  long long gcd(long long a, long long b) {
    if (a < b) swap(a, b);
    if (b == 0) return a;
    return gcd(b, a % b);
  }
  vector<long long> fact(N, 0ll);
  void factorial() {
    fact[0] = 1, fact[1] = 1;
    for (long long i = 2; i < N; i++) {
      fact[i] = (fact[i - 1] * i) % mod;
    }
  }
  long long ncr(long long n, long long r) {
    if (r > n) return 0;
    long long ans = fact[n] % mod;
    ans *= bPow(fact[r], mod - 2) % mod;
    ans %= mod;
    ans *= bPow(fact[n - r], mod - 2) % mod;
    ans %= mod;
    return ans;
  }
  vector<long long> primes(N, -1);
  void sieve() {
    iota(primes.begin(), primes.end(), 0);
    for (long long i = 2; i < N; ++i) {
      if (primes[i] == i) {
        for (long long j = i * 2; j < N; j += i) {
          primes[j] = i;
        }
      }
    }
  }
  long long add(long long a, long long b) {
    a %= mod, b %= mod;
    return ((a + b) % mod + mod) % mod;
  }
  long long mul(long long a, long long b) {
    a %= mod, b %= mod;
    return ((a * b) % mod + mod) % mod;
  }
  long long sub(long long a, long long b) {
    a %= mod, b %= mod;
    return ((a - b + 2 * mod) % mod + mod) % mod;
  }
  void solve() {
    long long n;
    cin >> n;
    vector<long long> a(n + 1);
    bool isPos = true;
    for (long long i = 1; i <= n; ++i) {
      cin >> a[i];
      if (i % 2 != a[i] % 2) {
        isPos = false;
      }
    }
    if (!isPos) {
      cout << -1 << "\n";
      return;
    }
    long long cur1 = n, r = n;
    vector<long long> ans;
    while (cur1 != 1) {
      long long cur2 = cur1 - 1, pos = n;
      for (long long i = 1; i <= r; ++i) {
        if (a[i] == cur1) {
          pos = i;
          break;
        }
      }
      ans.push_back(pos);
      reverse(a.begin() + 1, a.begin() + pos + 1);
      for (long long i = 1; i <= r; ++i) {
        if (a[i] == cur2) {
          pos = i;
          break;
        }
      }
      ans.push_back(pos - 1);
      reverse(a.begin() + 1, a.begin() + pos);
      ans.push_back(pos + 1);
      reverse(a.begin() + 1, a.begin() + pos + 2);
      ans.push_back(3);
      reverse(a.begin() + 1, a.begin() + 4);
      ans.push_back(r);
      reverse(a.begin() + 1, a.begin() + r + 1);
      cur1 -= 2, r -= 2;
    }
    cout << ans.size() << "\n";
    for (long long x : ans) cout << x << " ";
    cout << "\n";
  }
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    ;
    long long tc = 1, test = 1;
    cin >> tc;
    while (tc--) {
      solve();
    }
  }
  