#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline void read(T& x) {
  T t = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    t = (t << 3) + (t << 1) + ch - '0';
    ch = getchar();
  }
  x = f * t;
}
template <typename T, typename... Args>
inline void read(T& t, Args&... args) {
  read(t);
  read(args...);
}
const int inf = 0x3f3f3f3f;
const long long linf = 0x3f3f3f3f3f3f3f3f;
const double eps = 1e-8;
const int maxn = 3e3 + 9;
const long long mod = 998244353;
int n;
int a[maxn], pos[maxn];
void re(vector<int>& v, int p) {
  v.push_back(p);
  reverse(a + 1, a + p + 1);
  for (int i = 1; i <= p; ++i) pos[a[i]] = i;
}
void solve() {
  read(n);
  vector<int> ans;
  bool f = 0;
  for (int i = 1; i <= n; ++i) {
    read(a[i]);
    pos[a[i]] = i;
    if (i % 2 != a[i] % 2) {
      f = 1;
    }
  }
  if (f) {
    cout << -1 << '\n';
    return;
  }
  for (int i = n; i > 1; i -= 2) {
    int p1 = pos[i], p2 = pos[i - 1];
    if (p1 != 1) re(ans, p1);
    p1 = 1, p2 = pos[i - 1];
    if (p2 != 2) {
      re(ans, p2 - 1);
    }
    re(ans, i);
    p1 = pos[i], p2 = pos[i - 1];
    re(ans, p1);
    re(ans, i);
  }
  cout << ans.size() << '\n';
  for (auto i : ans) cout << i << ' ';
  cout << '\n';
}
signed main() {
  int T = 1;
  read(T);
  while (T--) {
    solve();
  }
  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;
template <typename T>
inline void read(T& x) {
  T t = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    t = (t << 3) + (t << 1) + ch - '0';
    ch = getchar();
  }
  x = f * t;
}
template <typename T, typename... Args>
inline void read(T& t, Args&... args) {
  read(t);
  read(args...);
}
const int inf = 0x3f3f3f3f;
const long long linf = 0x3f3f3f3f3f3f3f3f;
const double eps = 1e-8;
const int maxn = 3e3 + 9;
const long long mod = 998244353;
int n;
int a[maxn], pos[maxn];
void re(vector<int>& v, int p) {
  v.push_back(p);
  reverse(a + 1, a + p + 1);
  for (int i = 1; i <= p; ++i) pos[a[i]] = i;
}
void solve() {
  read(n);
  vector<int> ans;
  bool f = 0;
  for (int i = 1; i <= n; ++i) {
    read(a[i]);
    pos[a[i]] = i;
    if (i % 2 != a[i] % 2) {
      f = 1;
    }
  }
  if (f) {
    cout << -1 << '\n';
    return;
  }
  for (int i = n; i > 1; i -= 2) {
    int p1 = pos[i], p2 = pos[i - 1];
    if (p1 != 1) re(ans, p1);
    p1 = 1, p2 = pos[i - 1];
    if (p2 != 2) {
      re(ans, p2 - 1);
    }
    re(ans, i);
    p1 = pos[i], p2 = pos[i - 1];
    re(ans, p1);
    re(ans, i);
  }
  cout << ans.size() << '\n';
  for (auto i : ans) cout << i << ' ';
  cout << '\n';
}
signed main() {
  int T = 1;
  read(T);
  while (T--) {
    solve();
  }
  return 0;
}

```