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 M = 1e9 + 7;
  long long pmp(long long x, long long n) {
    long long a = 1;
    while (n > 0) {
      if (n & 1) a = (a * x) % M;
      x = (x * x) % M;
      n = n >> 1;
    }
    return a;
  }
  void solve() {
    long long n;
    cin >> n;
    vector<long long> a(n);
    vector<long long> inv(n);
    for (long long i = 0; i < n; i++) {
      cin >> a[i];
      --a[i];
      inv[a[i]] = i;
    }
    for (long long i = 0; i < n; i++) {
      if (i % 2 != a[i] % 2) {
        cout << "-1\n";
        return;
      }
    }
    vector<long long> op;
    long long nxt = n - 1;
    while (nxt) {
      long long last = inv[nxt];
      long long prev = inv[nxt - 1];
      if (last == nxt && prev == nxt - 1) {
        nxt -= 2;
        continue;
      }
      if (last > prev || last) {
        op.push_back(last + 1);
        reverse(a.begin(), a.begin() + last + 1);
        prev = (last > prev ? last - prev : prev);
        last = 0;
      }
      if (prev - 1) {
        op.push_back(prev);
        reverse(a.begin(), a.begin() + prev);
        last = prev - 1 - last;
      }
      if (last < prev) {
        op.push_back(prev + 2);
        reverse(a.begin(), a.begin() + prev + 2);
      }
      op.push_back(3);
      reverse(a.begin(), a.begin() + 3);
      op.push_back(nxt + 1);
      reverse(a.begin(), a.begin() + nxt + 1);
      nxt -= 2;
      for (long long i = 0; i < nxt + 1; i++) inv[a[i]] = i;
    }
    cout << op.size() << "\n";
    for (auto x : op) cout << x << " ";
    cout << "\n";
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    long long t;
    t = 1;
    cin >> t;
    while (t--) {
      solve();
    }
    cerr << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC
         << "ms\n";
    return 0;
  }
  