#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
const ld eps = 1e-8;
template <class U, class V>
ostream& operator<<(ostream& out, const pair<U, V>& u) {
  return out << "(" << u.first << ", " << u.second << ")";
}
template <class Con, class = decltype(begin(declval<Con>()))>
typename enable_if<!is_same<Con, string>::value, ostream&>::type operator<<(
    ostream& out, const Con& con) {
  out << "{";
  for (auto beg = con.begin(), it = beg; it != con.end(); ++it)
    out << (it == beg ? "" : ", ") << *it;
  return out << "}";
}
template <size_t i, class T>
ostream& print_tuple_utils(ostream& out, const T& tup) {
  if constexpr (i == tuple_size<T>::value)
    return out << ")";
  else
    return print_tuple_utils<i + 1, T>(out << (i ? ", " : "(") << get<i>(tup),
                                       tup);
}
template <class... U>
ostream& operator<<(ostream& out, const tuple<U...>& u) {
  return print_tuple_utils<0, tuple<U...>>(out, u);
}
const int maxn = 3000;
int n;
int a[maxn];
int pos_of(int x) {
  int ans = 0;
  while (ans < n and a[ans] != x) ++ans;
  assert(ans < n);
  return ans;
}
vector<int> ans;
void do_rev(int len) {
  assert(len & 1);
  ans.push_back(len);
  reverse(a, a + len);
}
int main() {
  cin.tie(0)->sync_with_stdio(0);
  int ntest;
  cin >> ntest;
  while (ntest--) {
    cin >> n;
    for (int i = (int)(0); i < (int)(n); ++i) cin >> a[i];
    ans.clear();
    bool ok = true;
    for (int len = n; len > 1; len -= 2) {
      int mx = len;
      int pmx = pos_of(mx);
      if (pmx % 2 == 1) {
        ok = false;
        break;
      }
      do_rev(pmx + 1);
      int pmx1 = pos_of(mx - 1);
      if (pmx1 % 2 == 0) {
        ok = false;
        break;
      }
      do_rev(pmx1);
      do_rev(pmx1 + 2);
      do_rev(3);
      do_rev(len);
    }
    if (!ok) {
      cout << "-1\n";
      continue;
    }
    cout << ((int)ans.size()) << '\n';
    for (auto i : ans) cout << i << ' ';
    cout << '\n';
    assert(is_sorted(a, a + n));
  }
  return 0;
}

### Prompt

Create a solution in cpp for the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
const ld eps = 1e-8;
template <class U, class V>
ostream& operator<<(ostream& out, const pair<U, V>& u) {
  return out << "(" << u.first << ", " << u.second << ")";
}
template <class Con, class = decltype(begin(declval<Con>()))>
typename enable_if<!is_same<Con, string>::value, ostream&>::type operator<<(
    ostream& out, const Con& con) {
  out << "{";
  for (auto beg = con.begin(), it = beg; it != con.end(); ++it)
    out << (it == beg ? "" : ", ") << *it;
  return out << "}";
}
template <size_t i, class T>
ostream& print_tuple_utils(ostream& out, const T& tup) {
  if constexpr (i == tuple_size<T>::value)
    return out << ")";
  else
    return print_tuple_utils<i + 1, T>(out << (i ? ", " : "(") << get<i>(tup),
                                       tup);
}
template <class... U>
ostream& operator<<(ostream& out, const tuple<U...>& u) {
  return print_tuple_utils<0, tuple<U...>>(out, u);
}
const int maxn = 3000;
int n;
int a[maxn];
int pos_of(int x) {
  int ans = 0;
  while (ans < n and a[ans] != x) ++ans;
  assert(ans < n);
  return ans;
}
vector<int> ans;
void do_rev(int len) {
  assert(len & 1);
  ans.push_back(len);
  reverse(a, a + len);
}
int main() {
  cin.tie(0)->sync_with_stdio(0);
  int ntest;
  cin >> ntest;
  while (ntest--) {
    cin >> n;
    for (int i = (int)(0); i < (int)(n); ++i) cin >> a[i];
    ans.clear();
    bool ok = true;
    for (int len = n; len > 1; len -= 2) {
      int mx = len;
      int pmx = pos_of(mx);
      if (pmx % 2 == 1) {
        ok = false;
        break;
      }
      do_rev(pmx + 1);
      int pmx1 = pos_of(mx - 1);
      if (pmx1 % 2 == 0) {
        ok = false;
        break;
      }
      do_rev(pmx1);
      do_rev(pmx1 + 2);
      do_rev(3);
      do_rev(len);
    }
    if (!ok) {
      cout << "-1\n";
      continue;
    }
    cout << ((int)ans.size()) << '\n';
    for (auto i : ans) cout << i << ' ';
    cout << '\n';
    assert(is_sorted(a, a + n));
  }
  return 0;
}

```