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;
  template <class T>
  bool chkmax(T& a, const T& b) {
    return a < b ? a = b, 1 : 0;
  }
  int read() {
    int num = 0, f = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9') {
      if (ch == '-') f = -1;
      ch = getchar();
    }
    while (ch >= '0' && ch <= '9') {
      num *= 10;
      num += ch - '0';
      ch = getchar();
    }
    return num * f;
  }
  long long read1() {
    long long num = 0, f = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9') {
      if (ch == '-') f = -1;
      ch = getchar();
    }
    while (ch >= '0' && ch <= '9') {
      num *= 10;
      num += ch - '0';
      ch = getchar();
    }
    return num * f;
  }
  int idx(char x) { return x - 'a'; }
  int a[1000010];
  int ans[4000010];
  long long n, m;
  int t, tt;
  int main() {
    t = read();
    while (t--) {
      n = read();
      for (int i = 1; i <= n; i++) {
        a[i] = read();
      }
      int tot = 0;
      bool flag = true;
      int cnt = 0;
      for (int k = n; k > 1; k -= 2) {
        int x = 1;
        while (a[x] != k) x++;
        int y = 1;
        while (a[y] != k - 1) y++;
        if ((y & 1) || (x % 2 == 0)) {
          flag = false;
          break;
        }
        reverse(a + 1, a + x + 1);
        ans[++tot] = x;
        if (y < x) {
          tt = x;
          x = tt + 1 - x;
          y = tt + 1 - y;
        }
        reverse(a + 1, a + y);
        ans[++tot] = y - 1;
        reverse(a + 1, a + y + 2);
        ans[++tot] = y + 1;
        reverse(a + 1, a + 4);
        ans[++tot] = 3;
        reverse(a + 1, a + k + 1);
        ans[++tot] = k;
      }
      if (a[1] != 1 || flag == false)
        cout << -1 << endl;
      else {
        cout << tot << endl;
        for (int i = 1; i <= tot; i++) cout << ans[i] << ' ';
        if (tot || cnt) cout << endl;
      }
    }
    return 0;
  }
  