#include <bits/stdc++.h>
using namespace std;
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
long long T, n, a[3000], x, y;
bool flag;
void doit(int x) {
  for (int i = 1; i <= x / 2; i++) swap(a[i], a[x - i + 1]);
}
int main() {
  T = read();
  while (T--) {
    n = read();
    flag = true;
    for (int i = 1; i <= n; i++) {
      a[i] = read();
      if (a[i] % 2 != i % 2) flag = false;
    }
    if (!flag) {
      cout << -1 << endl;
      continue;
    }
    cout << (n - 1) * 5 / 2 << endl;
    for (int i = n / 2; i >= 1; i--) {
      for (int j = 1; j <= n; j++) {
        if (a[j] == i * 2 + 1) {
          x = j;
          break;
        }
      }
      for (int j = 1; j <= n; j++) {
        if (a[j] == i * 2) {
          y = j;
          break;
        }
      }
      cout << x << " ";
      doit(x);
      if (y < x) y = x - y + 1;
      cout << y - 1 << " ";
      doit(y - 1);
      cout << y + 1 << " ";
      doit(y + 1);
      cout << 3 << " ";
      doit(3);
      cout << i * 2 + 1 << " ";
      doit(i * 2 + 1);
    }
    cout << endl;
  }
  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;
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
long long T, n, a[3000], x, y;
bool flag;
void doit(int x) {
  for (int i = 1; i <= x / 2; i++) swap(a[i], a[x - i + 1]);
}
int main() {
  T = read();
  while (T--) {
    n = read();
    flag = true;
    for (int i = 1; i <= n; i++) {
      a[i] = read();
      if (a[i] % 2 != i % 2) flag = false;
    }
    if (!flag) {
      cout << -1 << endl;
      continue;
    }
    cout << (n - 1) * 5 / 2 << endl;
    for (int i = n / 2; i >= 1; i--) {
      for (int j = 1; j <= n; j++) {
        if (a[j] == i * 2 + 1) {
          x = j;
          break;
        }
      }
      for (int j = 1; j <= n; j++) {
        if (a[j] == i * 2) {
          y = j;
          break;
        }
      }
      cout << x << " ";
      doit(x);
      if (y < x) y = x - y + 1;
      cout << y - 1 << " ";
      doit(y - 1);
      cout << y + 1 << " ";
      doit(y + 1);
      cout << 3 << " ";
      doit(3);
      cout << i * 2 + 1 << " ";
      doit(i * 2 + 1);
    }
    cout << endl;
  }
  return 0;
}

```