#include <bits/stdc++.h>
int T, n, f, a[2100], b[2100];
void rev(int x) {
  (x & 1) || (++x), printf("%d ", x);
  for (int i = 1, t; i <= (x >> 1); ++i)
    b[a[i]] = x - i + 1, b[a[x - i + 1]] = i, t = a[i], a[i] = a[x - i + 1],
    a[x - i + 1] = t;
}
int main() {
  scanf("%d", &T);
  while (T--) {
    scanf("%d", &n), f = 0;
    for (int i = 1; i <= n; ++i)
      scanf("%d", a + i), b[a[i]] = i, f |= (a[i] - i & 1);
    if (f) {
      puts("-1");
      continue;
    }
    printf("%d\n", (n >> 1) * 5);
    for (int i = n - 1; i; i -= 2)
      rev(b[i + 1]), rev(b[i] - 1), rev(b[i] + 1), rev(3), rev(i + 1);
    putchar('\n');
  }
  return 0;
}

### 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>
int T, n, f, a[2100], b[2100];
void rev(int x) {
  (x & 1) || (++x), printf("%d ", x);
  for (int i = 1, t; i <= (x >> 1); ++i)
    b[a[i]] = x - i + 1, b[a[x - i + 1]] = i, t = a[i], a[i] = a[x - i + 1],
    a[x - i + 1] = t;
}
int main() {
  scanf("%d", &T);
  while (T--) {
    scanf("%d", &n), f = 0;
    for (int i = 1; i <= n; ++i)
      scanf("%d", a + i), b[a[i]] = i, f |= (a[i] - i & 1);
    if (f) {
      puts("-1");
      continue;
    }
    printf("%d\n", (n >> 1) * 5);
    for (int i = n - 1; i; i -= 2)
      rev(b[i + 1]), rev(b[i] - 1), rev(b[i] + 1), rev(3), rev(i + 1);
    putchar('\n');
  }
  return 0;
}

```