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 mod = 1000000007, inf = 2e18;
  const int N = 200001;
  const double ep = 1e-10;
  void Darling(int kase) {
    int n;
    cin >> n;
    int a[n];
    for (int i = 0; (1 < 0 && i > n) || (1 > 0 && i < n); i += 1) cin >> a[i];
    for (int i = 0; (1 < 0 && i > n) || (1 > 0 && i < n); i += 1)
      if (a[i] % 2 == i % 2) {
        cout << -1 << '\n';
        return;
      }
    vector<int> z;
    int o = 1;
    while (o < n) {
      int i1 = 0, i2 = 1;
      for (int i = 0; (1 < 0 && i > n) || (1 > 0 && i < n); i += 1) {
        if (a[i] == o) i1 = i;
        if (a[i] == o + 1) i2 = i;
      }
      z.push_back(i1 + 1), reverse(a, a + i1 + 1);
      if (i2 < i1) i2 = i1 - i2;
      i1 = 0;
      z.push_back(i2), reverse(a, a + i2);
      i1 = i2 - 1;
      z.push_back(i2 + 2), reverse(a, a + i2 + 2);
      z.push_back(3), reverse(a, a + 3);
      z.push_back(n - o + 1), reverse(a, a + n - o + 1);
      o += 2;
    }
    cout << z.size() + 1 << '\n';
    for (int x : z) cout << x << ' ';
    cout << n << '\n';
  }
  int main() {
    ios_base::sync_with_stdio(0), cin.tie(0);
    int T = 1;
    cin >> T;
    for (int K = 0; K < T; K++) Darling(K + 1);
  }
  