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 int MAXN = 2105 * 5;
  int n;
  int a[MAXN];
  int ans[MAXN], anstot;
  long long Read() {
    long long x = 0, f = 1;
    char c = getchar();
    while (c > '9' || c < '0') {
      if (c == '-') f = -1;
      c = getchar();
    }
    while (c >= '0' && c <= '9') {
      x = (x * 10) + (c ^ 48);
      c = getchar();
    }
    return x * f;
  }
  template <typename T>
  void Put1(T x) {
    if (x > 9) Put1(x / 10);
    putchar(x % 10 ^ 48);
  }
  template <typename T>
  void Put(T x, char c = -1) {
    if (x < 0) putchar('-'), x = -x;
    Put1(x);
    if (c >= 0) putchar(c);
  }
  template <typename T>
  T Max(T x, T y) {
    return x > y ? x : y;
  }
  template <typename T>
  T Min(T x, T y) {
    return x < y ? x : y;
  }
  template <typename T>
  T Abs(T x) {
    return x < 0 ? -x : x;
  }
  void solve(int x) {
    if (x == 1) return;
    for (int i = 1; i <= x / 2; ++i) swap(a[i], a[x - i + 1]);
    ans[++anstot] = x;
  }
  int main() {
    for (int T = Read(); T; --T) {
      n = Read();
      bool ls = 0;
      for (int i = 1; i <= n; ++i) {
        a[i] = Read();
        if ((a[i] & 1) ^ (i & 1)) ls = 1;
      }
      if (ls) {
        Put(-1, '\n');
        continue;
      }
      for (int i = 1; i <= n; ++i)
        if (a[i] ^ i) ls = 1;
      if (!ls) {
        Put(0, '\n');
        continue;
      }
      anstot = 0;
      for (int i = 1; i < n; i += 2) {
        int ID1 = 0, ID2 = 0;
        for (int j = 1; j <= n; ++j)
          if (a[j] == i)
            ID1 = j;
          else if (a[j] == i + 1)
            ID2 = j;
        if (ID1 > ID2) ID2 = ID1 - ID2 + 1;
        solve(ID1);
        solve(ID2 - 1);
        ID1 = ID2 - 1;
        int val = ID2 + 1;
        ID1 = val - ID1 + 1;
        ID2 = val - ID2 + 1;
        solve(val);
        solve(ID1);
        ID1 = 1;
        ID2 = 2;
        solve(n - i + 1);
      }
      solve(n);
      Put(anstot, '\n');
      for (int i = 1; i <= anstot; ++i) Put(ans[i], ' ');
      putchar('\n');
    }
    return 0;
  }
  