#include <bits/stdc++.h>
using namespace std;
std::mt19937 rnd(
    (int)std::chrono::steady_clock::now().time_since_epoch().count());
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
const int MAXN = 2021;
int n;
int a[MAXN];
vector<int> ans;
int cur[MAXN];
void move(int len) {
  assert(len % 2 == 1);
  ans.push_back(len);
  reverse(cur, cur + len);
}
bool solve() {
  assert(n % 2 == 1);
  for (int i = (0); i < (n); ++i)
    if ((a[i] % 2) != (i % 2)) return false;
  ans.clear();
  for (int i = (0); i < (n); ++i) cur[i] = a[i];
  for (int x = n - 1; x >= 2; x -= 2) {
    int xpos = -1;
    for (int i = (0); i < (n); ++i)
      if (cur[i] == x) xpos = i;
    assert(xpos != -1 && xpos % 2 == 0);
    if (xpos != 0) move(xpos + 1), xpos = 0;
    int y = x - 1;
    int ypos = -1;
    for (int i = (0); i < (n); ++i)
      if (cur[i] == y) ypos = i;
    assert(ypos != -1 && ypos % 2 == 1);
    if (ypos != 1) move(ypos), xpos = ypos - 1;
    if (xpos != 0) {
      move(ypos + 2), ypos = 1, xpos = 2;
      move(3), xpos = 0, ypos = 1;
    }
    move(x + 1);
  }
  return true;
}
void run() {
  scanf("%d", &n);
  for (int i = (0); i < (n); ++i) scanf("%d", &a[i]), --a[i];
  if (!solve()) {
    printf("%d\n", -1);
    return;
  }
  printf("%d\n", ((int)(ans).size()));
  for (int i = (0); i < (((int)(ans).size())); ++i) {
    if (i != 0) printf(" ");
    printf("%d", ans[i]);
  }
  puts("");
}
int main() {
  int ncase;
  scanf("%d", &ncase);
  for (int i = (1); i <= (ncase); ++i) run();
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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;
std::mt19937 rnd(
    (int)std::chrono::steady_clock::now().time_since_epoch().count());
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
const int MAXN = 2021;
int n;
int a[MAXN];
vector<int> ans;
int cur[MAXN];
void move(int len) {
  assert(len % 2 == 1);
  ans.push_back(len);
  reverse(cur, cur + len);
}
bool solve() {
  assert(n % 2 == 1);
  for (int i = (0); i < (n); ++i)
    if ((a[i] % 2) != (i % 2)) return false;
  ans.clear();
  for (int i = (0); i < (n); ++i) cur[i] = a[i];
  for (int x = n - 1; x >= 2; x -= 2) {
    int xpos = -1;
    for (int i = (0); i < (n); ++i)
      if (cur[i] == x) xpos = i;
    assert(xpos != -1 && xpos % 2 == 0);
    if (xpos != 0) move(xpos + 1), xpos = 0;
    int y = x - 1;
    int ypos = -1;
    for (int i = (0); i < (n); ++i)
      if (cur[i] == y) ypos = i;
    assert(ypos != -1 && ypos % 2 == 1);
    if (ypos != 1) move(ypos), xpos = ypos - 1;
    if (xpos != 0) {
      move(ypos + 2), ypos = 1, xpos = 2;
      move(3), xpos = 0, ypos = 1;
    }
    move(x + 1);
  }
  return true;
}
void run() {
  scanf("%d", &n);
  for (int i = (0); i < (n); ++i) scanf("%d", &a[i]), --a[i];
  if (!solve()) {
    printf("%d\n", -1);
    return;
  }
  printf("%d\n", ((int)(ans).size()));
  for (int i = (0); i < (((int)(ans).size())); ++i) {
    if (i != 0) printf(" ");
    printf("%d", ans[i]);
  }
  puts("");
}
int main() {
  int ncase;
  scanf("%d", &ncase);
  for (int i = (1); i <= (ncase); ++i) run();
  return 0;
}

```