#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline bool chkmax(T &a, T b) {
  return a < b ? a = b, 1 : 0;
}
template <typename T>
inline bool chkmin(T &a, T b) {
  return b < a ? a = b, 1 : 0;
}
template <typename T>
inline T read() {
  T _ = 0, __ = 1;
  char c_ = getchar();
  for (; c_ < '0' || c_ > '9'; c_ = getchar())
    if (c_ == '-') __ = -1;
  for (; c_ >= '0' && c_ <= '9'; c_ = getchar())
    _ = (_ << 1) + (_ << 3) + (c_ ^ 48);
  return _ * __;
}
vector<int> ans;
int n, a[2333];
void rev(int x, int &x1, int &x2) {
  reverse(a + 1, a + 1 + x);
  ans.push_back(x);
  if (x1 <= x) x1 = x - x1 + 1;
  if (x2 <= x) x2 = x - x2 + 1;
}
int main() {
  int T = read<int>();
  while (T--) {
    scanf("%d", &n);
    int fl = 0;
    for (int i = (1), i_end_ = (n); i <= i_end_; ++i) {
      scanf("%d", a + i);
      if ((a[i] + i) % 2) {
        fl = 1;
      }
    }
    if (fl) {
      puts("-1");
      goto SIGN;
    }
    ans.clear();
    for (int x = n; x > 1; x -= 2) {
      int pn, pn1;
      for (int i = (1), i_end_ = (n); i <= i_end_; ++i)
        if (a[i] == x)
          pn = i;
        else if (a[i] == x - 1)
          pn1 = i;
      rev(pn, pn, pn1);
      rev(pn1 - 1, pn, pn1);
      rev(pn1 + 1, pn, pn1);
      rev(pn, pn, pn1);
      rev(x, pn, pn1);
    }
    printf("%d\n", n / 2 * 5);
    for (int i = 0, i_end_ = (n / 2 * 5); i < i_end_; ++i)
      printf("%d ", ans[i]);
    puts("");
  SIGN:;
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 bool chkmax(T &a, T b) {
  return a < b ? a = b, 1 : 0;
}
template <typename T>
inline bool chkmin(T &a, T b) {
  return b < a ? a = b, 1 : 0;
}
template <typename T>
inline T read() {
  T _ = 0, __ = 1;
  char c_ = getchar();
  for (; c_ < '0' || c_ > '9'; c_ = getchar())
    if (c_ == '-') __ = -1;
  for (; c_ >= '0' && c_ <= '9'; c_ = getchar())
    _ = (_ << 1) + (_ << 3) + (c_ ^ 48);
  return _ * __;
}
vector<int> ans;
int n, a[2333];
void rev(int x, int &x1, int &x2) {
  reverse(a + 1, a + 1 + x);
  ans.push_back(x);
  if (x1 <= x) x1 = x - x1 + 1;
  if (x2 <= x) x2 = x - x2 + 1;
}
int main() {
  int T = read<int>();
  while (T--) {
    scanf("%d", &n);
    int fl = 0;
    for (int i = (1), i_end_ = (n); i <= i_end_; ++i) {
      scanf("%d", a + i);
      if ((a[i] + i) % 2) {
        fl = 1;
      }
    }
    if (fl) {
      puts("-1");
      goto SIGN;
    }
    ans.clear();
    for (int x = n; x > 1; x -= 2) {
      int pn, pn1;
      for (int i = (1), i_end_ = (n); i <= i_end_; ++i)
        if (a[i] == x)
          pn = i;
        else if (a[i] == x - 1)
          pn1 = i;
      rev(pn, pn, pn1);
      rev(pn1 - 1, pn, pn1);
      rev(pn1 + 1, pn, pn1);
      rev(pn, pn, pn1);
      rev(x, pn, pn1);
    }
    printf("%d\n", n / 2 * 5);
    for (int i = 0, i_end_ = (n / 2 * 5); i < i_end_; ++i)
      printf("%d ", ans[i]);
    puts("");
  SIGN:;
  }
  return 0;
}

```