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>
  #pragma GCC optimize("Ofast")
  using namespace std;
  inline long long read() {
    long long x = 0, f = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9') {
      if (ch == '-') f = -1;
      ch = getchar();
    }
    while (ch >= '0' && ch <= '9') {
      x = x * 10 + ch - '0';
      ch = getchar();
    }
    return x * f;
  }
  void write(long long x) {
    if (x < 0) putchar('-'), x = -x;
    if (x >= 10) write(x / 10);
    putchar(x % 10 + '0');
  }
  const long long N = 1000010, mod = 1e9 + 7;
  char s[N];
  long long n, a[N], A, B, xianA, xianB;
  vector<long long> ans;
  long long find(long long x) {
    long long ans = 0;
    for (long long i = (long long)(1); i <= (long long)(n); ++i)
      if (a[i] == x) ans = i;
    return ans;
  }
  int main() {
    for (long long T = read(); T--;) {
      n = read();
      bool fl = 1;
      ans.clear();
      for (long long i = (long long)(1); i <= (long long)(n); ++i) {
        a[i] = read();
        fl &= (a[i] & 1) == (i & 1);
      }
      if (!fl)
        puts("-1");
      else {
        for (long long i = n; i >= 3; i -= 2) {
          long long p;
          p = find(i);
          if (p > 0 && p != 1) ans.push_back(p), reverse(a + 1, a + p + 1);
          p = find(i - 1) - 1;
          if (p > 0 && p != 1) ans.push_back(p), reverse(a + 1, a + p + 1);
          p = find(i - 1) + 1;
          if (p > 0 && p != 1) ans.push_back(p), reverse(a + 1, a + p + 1);
          p = find(i);
          if (p > 0 && p != 1) ans.push_back(p), reverse(a + 1, a + p + 1);
          ans.push_back(i);
          reverse(a + 1, a + i + 1);
        }
        cout << ans.size() << endl;
        for (long long i = (long long)(0); i < (long long)(ans.size()); ++i)
          write(ans[i]), putchar(' ');
        puts("");
      }
    }
  }
  