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;
  template <typename T>
  inline void read(T &x) {
    x = 0;
    char c = getchar();
    bool f = false;
    for (; !isdigit(c); c = getchar()) f |= c == '-';
    for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
    if (f) x = -x;
  }
  template <typename T, typename... Arg>
  inline void read(T &x, Arg &...args) {
    read(x);
    read(args...);
  }
  template <typename T>
  inline void write(T x) {
    if (x < 0) putchar('-'), x = -x;
    if (x >= 10) write(x / 10);
    putchar(x % 10 + '0');
  }
  const int N = 3000;
  int t, n, a[N];
  vector<int> ans;
  int find(int x) {
    for (int i = 1;; i++)
      if (a[i] == x) return i;
  }
  void rever(int x) {
    ans.push_back(x);
    reverse(a + 1, a + 1 + x);
  }
  signed main() {
    read(t);
    while (t--) {
      ans.clear();
      read(n);
      for (int i = 1; i <= n; i++) read(a[i]);
      for (int i = 1; i <= n; i++)
        if (abs(a[i] - i) & 1) {
          puts("-1");
          goto loop;
        }
      for (int i = 1; i <= n / 2; i++) {
        int pos1 = find(2 * i - 1);
        rever(pos1);
        int pos2 = find(i * 2);
        rever(pos2 - 1);
        rever(pos2 + 1);
        rever(3);
        rever(n - 2 * i + 2);
      }
      rever(n);
      write(ans.size());
      putchar('\n');
      for (auto x : ans) write(x), putchar(' ');
      putchar('\n');
    loop:;
    }
  }
  