#include <bits/stdc++.h>
using namespace std;
const int N = 100005;
const int inf = 1 << 30;
const long long inff = 1ll << 60;
inline int read() {
  int x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = (x << 1) + (x << 3) + (c ^ 48);
    c = getchar();
  }
  return x * f;
}
int n, a[N], ans[N], cnt;
void xyizd_is_stupid(int pre) {
  for (int i = 1, j = pre; i < j; i++, j--) a[i] ^= a[j] ^= a[i] ^= a[j];
  ans[++cnt] = pre;
}
void solve() {
  n = read();
  for (int i = 1; i <= n; i++) a[i] = read();
  for (int i = 1; i <= n; i++)
    if ((a[i] + i) % 2) {
      puts("-1");
      return;
    }
  cnt = 0;
  while (n >= 3) {
    int pos_n, pos_m;
    for (int i = 1; i <= n; i++)
      if (a[i] == n) pos_n = i;
    xyizd_is_stupid(pos_n);
    for (int i = 1; i <= n; i++)
      if (a[i] == n - 1) pos_m = i;
    xyizd_is_stupid(pos_m - 1);
    xyizd_is_stupid(pos_m + 1);
    xyizd_is_stupid(3);
    xyizd_is_stupid(n);
    n -= 2;
  }
  printf("%d\n", cnt);
  for (int i = 1; i <= cnt; i++) printf("%d ", ans[i]);
  puts("");
}
int main() {
  int tests = 1;
  tests = read();
  while (tests--) {
    solve();
  }
  return 0;
}

### Prompt

Generate 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;
const int N = 100005;
const int inf = 1 << 30;
const long long inff = 1ll << 60;
inline int read() {
  int x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = (x << 1) + (x << 3) + (c ^ 48);
    c = getchar();
  }
  return x * f;
}
int n, a[N], ans[N], cnt;
void xyizd_is_stupid(int pre) {
  for (int i = 1, j = pre; i < j; i++, j--) a[i] ^= a[j] ^= a[i] ^= a[j];
  ans[++cnt] = pre;
}
void solve() {
  n = read();
  for (int i = 1; i <= n; i++) a[i] = read();
  for (int i = 1; i <= n; i++)
    if ((a[i] + i) % 2) {
      puts("-1");
      return;
    }
  cnt = 0;
  while (n >= 3) {
    int pos_n, pos_m;
    for (int i = 1; i <= n; i++)
      if (a[i] == n) pos_n = i;
    xyizd_is_stupid(pos_n);
    for (int i = 1; i <= n; i++)
      if (a[i] == n - 1) pos_m = i;
    xyizd_is_stupid(pos_m - 1);
    xyizd_is_stupid(pos_m + 1);
    xyizd_is_stupid(3);
    xyizd_is_stupid(n);
    n -= 2;
  }
  printf("%d\n", cnt);
  for (int i = 1; i <= cnt; i++) printf("%d ", ans[i]);
  puts("");
}
int main() {
  int tests = 1;
  tests = read();
  while (tests--) {
    solve();
  }
  return 0;
}

```