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;
  inline char gc() {
    static char buf[100000], *p1 = buf, *p2 = buf;
    return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2)
               ? EOF
               : *p1++;
  }
  inline long long read() {
    char c = getchar();
    long long su = 0, f = 1;
    for (; c < '0' || c > '9'; c = getchar())
      if (c == '-') f = -1;
    for (; c >= '0' && c <= '9'; c = getchar()) su = su * 10 + c - '0';
    return su * f;
  }
  inline void write(long long x) {
    if (x < 0) {
      putchar('-');
      write(-x);
      return;
    }
    if (x >= 10) write(x / 10);
    putchar(x % 10 + '0');
  }
  inline void writesp(long long x) { write(x), putchar(' '); }
  inline void writeln(long long x) {
    write(x);
    putchar('\n');
  }
  int T, n, a[400005], bl;
  vector<int> ans;
  signed main() {
    T = read();
    while (T--) {
      n = read();
      bl = 0;
      for (int i = 1; i <= n; i++) {
        a[i] = read();
        if (a[i] % 2 != i % 2) bl = 1;
      }
      if (bl)
        puts("-1");
      else {
        for (int i = n; i >= 3; i -= 2) {
          int x = 0, y = 0;
          for (int j = 1; j <= i; j++)
            if (a[j] == i) {
              x = j;
            }
          for (int j = 1; j <= i; j++)
            if (a[j] == i - 1) {
              y = j;
            }
          ans.push_back(x);
          reverse(a + 1, a + x + 1);
          for (int j = 1; j <= i; j++)
            if (a[j] == i - 1) {
              y = j;
            }
          ans.push_back(y - 1);
          reverse(a + 1, a + y - 1 + 1);
          ans.push_back(i);
          reverse(a + 1, a + i + 1);
          for (int j = 1; j <= i; j++)
            if (a[j] == i) {
              x = j;
            }
          ans.push_back(x);
          reverse(a + 1, a + x + 1);
          ans.push_back(i);
          reverse(a + 1, a + i + 1);
        }
        for (int i = 1; i <= n; i++) {
          if (a[i] != i) bl = 1;
        }
        if (bl)
          puts("-1");
        else {
          writeln(ans.size());
          for (auto u : ans) writesp(u);
          puts("");
        }
      }
      ans.clear();
    }
  }
  