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 N1 = 2030;
  const long long inf = 0x3f3f3f3f3f3fll;
  template <typename T>
  inline void read(T &x) {
    x = 0;
    T k = 1;
    char c = getchar();
    while (!isdigit(c)) {
      if (c == '-') k = -1;
      c = getchar();
    }
    while (isdigit(c)) {
      x = x * 10 + c - '0';
      c = getchar();
    }
    x *= k;
  }
  int n, p, cnt;
  int a[N1], ans[N1 * 3];
  int find(int w) {
    for (int i = 1; i <= n; i++)
      if (a[i] == w) return i;
    return -1;
  }
  void revers(int x) {
    ans[++cnt] = x;
    for (int i = 1; i < x - i + 1; i++) swap(a[i], a[x - i + 1]);
  }
  void solve() {
    read(n);
    cnt = 0;
    for (int i = 1; i <= n; i++) read(a[i]);
    for (int i = 1; i <= n; i++)
      if ((a[i] & 1) ^ (i & 1)) {
        puts("-1");
        return;
      }
    for (int i = n; i >= 3; i -= 2) {
      int x = find(i - 1), y = find(i);
      if (x <= y) {
        revers(y);
        x = find(i - 1);
        revers(x - 1);
        revers(i);
        y = find(i);
        revers(y);
        revers(i);
      } else {
        revers(y);
        revers(x - 1);
        revers(i);
        y = find(i);
        revers(y);
        revers(i);
      }
    }
    printf("%d\n", cnt);
    for (int i = 1; i <= cnt; i++) {
      printf("%d", ans[i]);
      if (i == cnt)
        printf("\n");
      else
        printf(" ");
    }
  }
  int main() {
    int Te;
    read(Te);
    while (Te--) solve();
    return 0;
  }
  