#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline void read(T &x) {
  x = 0;
  char c = getchar();
  bool f = false;
  for (; !isdigit(c); c = getchar()) f |= c == '-';
  for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
  if (f) x = -x;
}
template <typename T, typename... Arg>
inline void read(T &x, Arg &...args) {
  read(x);
  read(args...);
}
template <typename T>
inline void write(T x) {
  if (x < 0) putchar('-'), x = -x;
  if (x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}
const int N = 3000;
int t, n, a[N];
vector<int> ans;
int find(int x) {
  for (int i = 1;; i++)
    if (a[i] == x) return i;
}
void rever(int x) {
  ans.push_back(x);
  reverse(a + 1, a + 1 + x);
}
signed main() {
  read(t);
  while (t--) {
    ans.clear();
    read(n);
    for (int i = 1; i <= n; i++) read(a[i]);
    for (int i = 1; i <= n; i++)
      if (abs(a[i] - i) & 1) {
        puts("-1");
        goto loop;
      }
    for (int i = 1; i <= n / 2; i++) {
      int pos1 = find(2 * i - 1);
      rever(pos1);
      int pos2 = find(i * 2);
      rever(pos2 - 1);
      rever(pos2 + 1);
      rever(3);
      rever(n - 2 * i + 2);
    }
    rever(n);
    write(ans.size());
    putchar('\n');
    for (auto x : ans) write(x), putchar(' ');
    putchar('\n');
  loop:;
  }
}

### Prompt

Please formulate a Cpp solution to 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) {
  x = 0;
  char c = getchar();
  bool f = false;
  for (; !isdigit(c); c = getchar()) f |= c == '-';
  for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
  if (f) x = -x;
}
template <typename T, typename... Arg>
inline void read(T &x, Arg &...args) {
  read(x);
  read(args...);
}
template <typename T>
inline void write(T x) {
  if (x < 0) putchar('-'), x = -x;
  if (x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}
const int N = 3000;
int t, n, a[N];
vector<int> ans;
int find(int x) {
  for (int i = 1;; i++)
    if (a[i] == x) return i;
}
void rever(int x) {
  ans.push_back(x);
  reverse(a + 1, a + 1 + x);
}
signed main() {
  read(t);
  while (t--) {
    ans.clear();
    read(n);
    for (int i = 1; i <= n; i++) read(a[i]);
    for (int i = 1; i <= n; i++)
      if (abs(a[i] - i) & 1) {
        puts("-1");
        goto loop;
      }
    for (int i = 1; i <= n / 2; i++) {
      int pos1 = find(2 * i - 1);
      rever(pos1);
      int pos2 = find(i * 2);
      rever(pos2 - 1);
      rever(pos2 + 1);
      rever(3);
      rever(n - 2 * i + 2);
    }
    rever(n);
    write(ans.size());
    putchar('\n');
    for (auto x : ans) write(x), putchar(' ');
    putchar('\n');
  loop:;
  }
}

```