#include <bits/stdc++.h>
#pragma GCC optimize "Ofast"
using namespace std;
int t, n, a[2100];
vector<int> ans;
void rev(int x) {
  reverse(a + 1, a + x + 1);
  ans.push_back(x);
}
bool check() {
  int ret = 0;
  vector<int> tmp;
  for (int i = 0; i < ans.size(); i++) {
    if (i != ans.size() - 1 and ans[i] == ans[i + 1]) {
      i++;
      ret = 1;
    } else {
      tmp.push_back(ans[i]);
    }
  }
  swap(ans, tmp);
  return ret;
}
int main() {
  cin >> t;
  while (t--) {
    bool imp = 0;
    cin >> n;
    ans.clear();
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
      if (a[i] % 2 != i % 2) imp = 1;
    }
    if (imp) {
      cout << "-1\n";
      continue;
    }
    for (int i = n; i >= 3; i -= 2) {
      int pos = 0, pos2 = 0;
      for (int j = 1; j <= n; j++) {
        if (a[j] == i) pos = j;
      }
      for (int j = 1; j <= n; j++) {
        if (a[j] == i - 1) pos2 = j;
      }
      int pos3 = 0;
      for (int j = 1; j <= n; j++) {
        if (a[j] == i) pos3 = j;
      }
      rev(pos3);
      int pos4 = 0;
      for (int j = 1; j <= n; j++) {
        if (a[j] == i - 1) pos4 = j;
      }
      rev(pos4 - 1);
      rev(pos4 + 1);
      rev(3);
      rev(i);
    }
    for (int i = 1; i <= n; i++) {
      if (a[i] != i) {
        imp = 1;
        break;
      }
    }
    if (imp)
      cout << "-1\n";
    else {
      while (check()) {
      }
      if (ans.size() > (5 * n) / 2)
        cout << "-1\n";
      else {
        cout << ans.size() << '\n';
        for (int i = 0; i < ans.size(); i++)
          cout << ans[i] << " \n"[i == ans.size() - 1];
      }
    }
  }
}

### Prompt

Develop a solution in Cpp 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>
#pragma GCC optimize "Ofast"
using namespace std;
int t, n, a[2100];
vector<int> ans;
void rev(int x) {
  reverse(a + 1, a + x + 1);
  ans.push_back(x);
}
bool check() {
  int ret = 0;
  vector<int> tmp;
  for (int i = 0; i < ans.size(); i++) {
    if (i != ans.size() - 1 and ans[i] == ans[i + 1]) {
      i++;
      ret = 1;
    } else {
      tmp.push_back(ans[i]);
    }
  }
  swap(ans, tmp);
  return ret;
}
int main() {
  cin >> t;
  while (t--) {
    bool imp = 0;
    cin >> n;
    ans.clear();
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
      if (a[i] % 2 != i % 2) imp = 1;
    }
    if (imp) {
      cout << "-1\n";
      continue;
    }
    for (int i = n; i >= 3; i -= 2) {
      int pos = 0, pos2 = 0;
      for (int j = 1; j <= n; j++) {
        if (a[j] == i) pos = j;
      }
      for (int j = 1; j <= n; j++) {
        if (a[j] == i - 1) pos2 = j;
      }
      int pos3 = 0;
      for (int j = 1; j <= n; j++) {
        if (a[j] == i) pos3 = j;
      }
      rev(pos3);
      int pos4 = 0;
      for (int j = 1; j <= n; j++) {
        if (a[j] == i - 1) pos4 = j;
      }
      rev(pos4 - 1);
      rev(pos4 + 1);
      rev(3);
      rev(i);
    }
    for (int i = 1; i <= n; i++) {
      if (a[i] != i) {
        imp = 1;
        break;
      }
    }
    if (imp)
      cout << "-1\n";
    else {
      while (check()) {
      }
      if (ans.size() > (5 * n) / 2)
        cout << "-1\n";
      else {
        cout << ans.size() << '\n';
        for (int i = 0; i < ans.size(); i++)
          cout << ans[i] << " \n"[i == ans.size() - 1];
      }
    }
  }
}

```