#include <bits/stdc++.h>
using namespace std;
const int N1 = 2030;
const long long inf = 0x3f3f3f3f3f3fll;
template <typename T>
inline void read(T &x) {
  x = 0;
  T k = 1;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') k = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    x = x * 10 + c - '0';
    c = getchar();
  }
  x *= k;
}
int n, p, cnt;
int a[N1], ans[N1 * 3];
int find(int w) {
  for (int i = 1; i <= n; i++)
    if (a[i] == w) return i;
  return -1;
}
void revers(int x) {
  ans[++cnt] = x;
  for (int i = 1; i < x - i + 1; i++) swap(a[i], a[x - i + 1]);
}
void solve() {
  read(n);
  cnt = 0;
  for (int i = 1; i <= n; i++) read(a[i]);
  for (int i = 1; i <= n; i++)
    if ((a[i] & 1) ^ (i & 1)) {
      puts("-1");
      return;
    }
  for (int i = n; i >= 3; i -= 2) {
    int x = find(i - 1), y = find(i);
    if (x <= y) {
      revers(y);
      x = find(i - 1);
      revers(x - 1);
      revers(i);
      y = find(i);
      revers(y);
      revers(i);
    } else {
      revers(y);
      revers(x - 1);
      revers(i);
      y = find(i);
      revers(y);
      revers(i);
    }
  }
  printf("%d\n", cnt);
  for (int i = 1; i <= cnt; i++) {
    printf("%d", ans[i]);
    if (i == cnt)
      printf("\n");
    else
      printf(" ");
  }
}
int main() {
  int Te;
  read(Te);
  while (Te--) 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 N1 = 2030;
const long long inf = 0x3f3f3f3f3f3fll;
template <typename T>
inline void read(T &x) {
  x = 0;
  T k = 1;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') k = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    x = x * 10 + c - '0';
    c = getchar();
  }
  x *= k;
}
int n, p, cnt;
int a[N1], ans[N1 * 3];
int find(int w) {
  for (int i = 1; i <= n; i++)
    if (a[i] == w) return i;
  return -1;
}
void revers(int x) {
  ans[++cnt] = x;
  for (int i = 1; i < x - i + 1; i++) swap(a[i], a[x - i + 1]);
}
void solve() {
  read(n);
  cnt = 0;
  for (int i = 1; i <= n; i++) read(a[i]);
  for (int i = 1; i <= n; i++)
    if ((a[i] & 1) ^ (i & 1)) {
      puts("-1");
      return;
    }
  for (int i = n; i >= 3; i -= 2) {
    int x = find(i - 1), y = find(i);
    if (x <= y) {
      revers(y);
      x = find(i - 1);
      revers(x - 1);
      revers(i);
      y = find(i);
      revers(y);
      revers(i);
    } else {
      revers(y);
      revers(x - 1);
      revers(i);
      y = find(i);
      revers(y);
      revers(i);
    }
  }
  printf("%d\n", cnt);
  for (int i = 1; i <= cnt; i++) {
    printf("%d", ans[i]);
    if (i == cnt)
      printf("\n");
    else
      printf(" ");
  }
}
int main() {
  int Te;
  read(Te);
  while (Te--) solve();
  return 0;
}

```