#include <bits/stdc++.h>
using namespace std;
const int MAXN = 2100;
int n;
int p[MAXN];
vector<int> ans;
void rev(int x) {
  if (x == 1) return;
  if (ans.empty() == false && ans.back() == x) {
    ans.pop_back();
    return;
  }
  ans.push_back(x);
  reverse(p + 1, p + x + 1);
}
void rev(int x, int &ind, int &ind1) {
  rev(x);
  if (ind <= x) ind = x - ind + 1;
  if (ind1 <= x) ind1 = x - ind1 + 1;
}
void solveTestcase() {
  ans.clear();
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> p[i];
  for (int i = 1; i <= n; i++) {
    if (p[i] % 2 != i % 2) {
      cout << "-1" << '\n';
      return;
    }
  }
  for (int num = n; num > 1; num -= 2) {
    if (p[num] == num && p[num - 1] == num - 1) continue;
    if (p[1] == num && p[2] == num - 1) {
      rev(num);
      continue;
    }
    int ind = -1, ind1 = -1;
    for (int i = 1; i <= num; i++) {
      if (p[i] == num) {
        ind = i;
      }
      if (p[i] == num - 1) {
        ind1 = i;
      }
    }
    if (p[ind - 1] == num - 1) {
      rev(ind);
      rev(num);
      continue;
    }
    if (p[ind + 1] == num - 1) {
      rev(num);
      ind = num - ind + 1;
      rev(ind);
      rev(num);
      continue;
    }
    if (ind < ind1) {
      rev(ind, ind, ind1);
      rev(ind1 - 1, ind, ind1);
      rev(num, ind, ind1);
      rev(ind);
      rev(num);
      continue;
    }
    if (ind > ind1) {
      rev(ind, ind, ind1);
      rev(ind1 - 1, ind, ind1);
      rev(num, ind, ind1);
      rev(ind, ind, ind1);
      rev(num, ind, ind1);
      continue;
    }
    cout << "-1" << '\n';
    return;
  }
  if (ans.size() > n * 2.5) {
    cout << "-1" << '\n';
    return;
  }
  cout << ans.size() << '\n';
  for (int x : ans) cout << x << " ";
  cout << '\n';
}
int main() {
  int T;
  cin >> T;
  while (T--) solveTestcase();
}

### Prompt

Please formulate 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 MAXN = 2100;
int n;
int p[MAXN];
vector<int> ans;
void rev(int x) {
  if (x == 1) return;
  if (ans.empty() == false && ans.back() == x) {
    ans.pop_back();
    return;
  }
  ans.push_back(x);
  reverse(p + 1, p + x + 1);
}
void rev(int x, int &ind, int &ind1) {
  rev(x);
  if (ind <= x) ind = x - ind + 1;
  if (ind1 <= x) ind1 = x - ind1 + 1;
}
void solveTestcase() {
  ans.clear();
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> p[i];
  for (int i = 1; i <= n; i++) {
    if (p[i] % 2 != i % 2) {
      cout << "-1" << '\n';
      return;
    }
  }
  for (int num = n; num > 1; num -= 2) {
    if (p[num] == num && p[num - 1] == num - 1) continue;
    if (p[1] == num && p[2] == num - 1) {
      rev(num);
      continue;
    }
    int ind = -1, ind1 = -1;
    for (int i = 1; i <= num; i++) {
      if (p[i] == num) {
        ind = i;
      }
      if (p[i] == num - 1) {
        ind1 = i;
      }
    }
    if (p[ind - 1] == num - 1) {
      rev(ind);
      rev(num);
      continue;
    }
    if (p[ind + 1] == num - 1) {
      rev(num);
      ind = num - ind + 1;
      rev(ind);
      rev(num);
      continue;
    }
    if (ind < ind1) {
      rev(ind, ind, ind1);
      rev(ind1 - 1, ind, ind1);
      rev(num, ind, ind1);
      rev(ind);
      rev(num);
      continue;
    }
    if (ind > ind1) {
      rev(ind, ind, ind1);
      rev(ind1 - 1, ind, ind1);
      rev(num, ind, ind1);
      rev(ind, ind, ind1);
      rev(num, ind, ind1);
      continue;
    }
    cout << "-1" << '\n';
    return;
  }
  if (ans.size() > n * 2.5) {
    cout << "-1" << '\n';
    return;
  }
  cout << ans.size() << '\n';
  for (int x : ans) cout << x << " ";
  cout << '\n';
}
int main() {
  int T;
  cin >> T;
  while (T--) solveTestcase();
}

```