#include <bits/stdc++.h>
using namespace std;
int read() {
  int tot = 0, fh = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') fh = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    tot = tot * 10 + c - '0';
    c = getchar();
  }
  return tot * fh;
}
void getstr(char *a) {
  char c = getchar();
  int len = 0;
  while (!((c >= 'A' && c <= 'Z') || (c >= 'a' && c <= 'z') ||
           (c >= '0' && c <= '9')))
    c = getchar();
  while ((c >= 'A' && c <= 'Z') || (c >= 'a' && c <= 'z') ||
         (c >= '0' && c <= '9')) {
    a[++len] = c;
    c = getchar();
  }
}
const int maxn = 4050;
int t, n, a[maxn], ans, tmp, f[maxn * 3];
void solve(int x) {
  for (int i = 1; i <= x / 2; i++) swap(a[i], a[x - i + 1]);
  ans++;
  f[ans] = x;
}
int main() {
  t = read();
  while (t--) {
    n = read();
    for (int i = 1; i <= n; i++) a[i] = read();
    ans = 0;
    bool flag = true;
    for (int i = 1; i <= n; i++)
      if ((a[i] % 2) != (i % 2)) {
        flag = false;
        break;
      }
    if (!flag) {
      cout << -1 << endl;
      continue;
    }
    for (int i = n; i >= 3; i -= 2) {
      if (a[i - 1] == (i - 1) && a[i] == i) continue;
      for (int j = 1; j <= i; j++)
        if (a[j] == i) tmp = j;
      if (tmp != 1) solve(tmp);
      for (int j = 1; j <= i; j++)
        if (a[j] == (i - 1)) tmp = j;
      if (tmp != 2) solve(tmp - 1);
      solve(i);
      for (int j = 1; j <= i; j++)
        if (a[j] == i) tmp = j;
      if (tmp != 1) solve(tmp);
      solve(i);
    }
    cout << ans << endl;
    for (int i = 1; i <= ans; i++) cout << f[i] << " ";
    if (ans != 0) 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;
int read() {
  int tot = 0, fh = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') fh = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    tot = tot * 10 + c - '0';
    c = getchar();
  }
  return tot * fh;
}
void getstr(char *a) {
  char c = getchar();
  int len = 0;
  while (!((c >= 'A' && c <= 'Z') || (c >= 'a' && c <= 'z') ||
           (c >= '0' && c <= '9')))
    c = getchar();
  while ((c >= 'A' && c <= 'Z') || (c >= 'a' && c <= 'z') ||
         (c >= '0' && c <= '9')) {
    a[++len] = c;
    c = getchar();
  }
}
const int maxn = 4050;
int t, n, a[maxn], ans, tmp, f[maxn * 3];
void solve(int x) {
  for (int i = 1; i <= x / 2; i++) swap(a[i], a[x - i + 1]);
  ans++;
  f[ans] = x;
}
int main() {
  t = read();
  while (t--) {
    n = read();
    for (int i = 1; i <= n; i++) a[i] = read();
    ans = 0;
    bool flag = true;
    for (int i = 1; i <= n; i++)
      if ((a[i] % 2) != (i % 2)) {
        flag = false;
        break;
      }
    if (!flag) {
      cout << -1 << endl;
      continue;
    }
    for (int i = n; i >= 3; i -= 2) {
      if (a[i - 1] == (i - 1) && a[i] == i) continue;
      for (int j = 1; j <= i; j++)
        if (a[j] == i) tmp = j;
      if (tmp != 1) solve(tmp);
      for (int j = 1; j <= i; j++)
        if (a[j] == (i - 1)) tmp = j;
      if (tmp != 2) solve(tmp - 1);
      solve(i);
      for (int j = 1; j <= i; j++)
        if (a[j] == i) tmp = j;
      if (tmp != 1) solve(tmp);
      solve(i);
    }
    cout << ans << endl;
    for (int i = 1; i <= ans; i++) cout << f[i] << " ";
    if (ans != 0) cout << endl;
  }
  return 0;
}

```