#include <bits/stdc++.h>
using namespace std;
const int MAXN = 2015;
int a[MAXN];
int ans[MAXN * 5], cnt;
int n;
void reserve(int x, int y) {
  while (x < y) swap(a[x++], a[y--]);
}
void deal(int x) {
  if (a[x] == x && a[x - 1] == x - 1) return;
  for (int i = 1; i <= x; i += 2) {
    if (a[i] != x) continue;
    ans[cnt++] = i;
    reserve(1, i);
    for (int j = 2; j <= x; j += 2) {
      if (a[j] == x - 1) {
        ans[cnt++] = j - 1;
        reserve(1, j - 1);
        ans[cnt++] = j + 1;
        reserve(1, j + 1);
        ans[cnt++] = 3;
        reserve(1, 3);
        ans[cnt++] = x;
        reserve(1, x);
        break;
      }
    }
    break;
  }
}
void solve() {
  cnt = 0;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++)
    if ((a[i] & 1) != (i & 1)) {
      printf("-1\n");
      return;
    }
  for (int i = n; i >= 3; i -= 2) deal(i);
  printf("%d\n", cnt);
  for (int i = 0; i < cnt; i++)
    printf("%d%c", ans[i], i == cnt - 1 ? '\n' : ' ');
}
int main() {
  int T = 1;
  scanf("%d", &T);
  for (int i = 1; i <= T; i++) 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;
const int MAXN = 2015;
int a[MAXN];
int ans[MAXN * 5], cnt;
int n;
void reserve(int x, int y) {
  while (x < y) swap(a[x++], a[y--]);
}
void deal(int x) {
  if (a[x] == x && a[x - 1] == x - 1) return;
  for (int i = 1; i <= x; i += 2) {
    if (a[i] != x) continue;
    ans[cnt++] = i;
    reserve(1, i);
    for (int j = 2; j <= x; j += 2) {
      if (a[j] == x - 1) {
        ans[cnt++] = j - 1;
        reserve(1, j - 1);
        ans[cnt++] = j + 1;
        reserve(1, j + 1);
        ans[cnt++] = 3;
        reserve(1, 3);
        ans[cnt++] = x;
        reserve(1, x);
        break;
      }
    }
    break;
  }
}
void solve() {
  cnt = 0;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++)
    if ((a[i] & 1) != (i & 1)) {
      printf("-1\n");
      return;
    }
  for (int i = n; i >= 3; i -= 2) deal(i);
  printf("%d\n", cnt);
  for (int i = 0; i < cnt; i++)
    printf("%d%c", ans[i], i == cnt - 1 ? '\n' : ' ');
}
int main() {
  int T = 1;
  scanf("%d", &T);
  for (int i = 1; i <= T; i++) solve();
  return 0;
}

```