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;
  void cppio() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
  }
  const int maxn = int(2e5) + 10;
  int n;
  int a[maxn];
  int ans[maxn], an;
  void Swap(int r) {
    int l = 1;
    if (l == r) return;
    ans[an++] = r;
    reverse(a + l, a + r + 1);
  }
  void Work() {
    cin >> n;
    for (int i = 1; i <= (n); ++i) cin >> a[i];
    for (int i = 1; i <= (n); ++i)
      if ((i & 1) != (a[i] & 1)) {
        cout << "-1\n";
        return;
      }
    an = 0;
    for (int z = n; 1 < z; z -= 2) {
      int zp = find(a + 1, a + z + 1, z) - a;
      int z2p = find(a + 2, a + z, z - 1) - a;
      if (zp < z2p) {
        Swap(zp);
        Swap(z2p - 1);
        Swap(z2p + 1);
        Swap(3);
        Swap(z);
      } else {
        Swap(zp);
        z2p = 1 + zp - z2p;
        Swap(z2p - 1);
        Swap(z2p + 1);
        Swap(3);
        Swap(z);
      }
    }
    cout << an << '\n';
    for (int i = 0; i < (an); ++i) cout << ans[i] << " \n"[i + 1 == an];
  }
  int main() {
    cppio();
    int tc;
    cin >> tc;
    for (int tci = 1; tci <= (tc); ++tci) {
      Work();
    }
    return 0;
  }
  