#include <bits/stdc++.h>
using namespace std;
const long long inf = 4e18;
const long long N = 2e5 + 10;
void testCase() {
  long long n;
  cin >> n;
  vector<long long> a(n + 1);
  for (long long i = 1; i <= n; i++) cin >> a[i];
  for (long long i = 1; i <= n; i++) {
    if ((a[i] % 2) != (i % 2)) {
      cout << -1 << '\n';
      return;
    }
  }
  auto find = [&](long long x) {
    for (long long i = 1; i <= n; i++) {
      if (a[i] == x) return i;
    }
    return x;
  };
  vector<long long> ans;
  auto doIt = [&](long long p) {
    assert(p & 1);
    ans.push_back(p);
    reverse(a.begin() + 1, a.begin() + p + 1);
  };
  for (long long r = n; r > 1; r -= 2) {
    long long p;
    p = find(r);
    doIt(p);
    p = find(r - 1) - 1;
    doIt(p);
    p = find(r - 1) + 1;
    doIt(p);
    p = 3;
    doIt(p);
    p = r;
    doIt(p);
  }
  assert(is_sorted(begin(a), end(a)));
  cout << (long long)ans.size() << '\n';
  for (long long &i : ans) cout << i << ' ';
  cout << '\n';
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long t_c = 1;
  cin >> t_c;
  while (t_c--) testCase();
}

### Prompt

In Cpp, your task is to solve 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 long long inf = 4e18;
const long long N = 2e5 + 10;
void testCase() {
  long long n;
  cin >> n;
  vector<long long> a(n + 1);
  for (long long i = 1; i <= n; i++) cin >> a[i];
  for (long long i = 1; i <= n; i++) {
    if ((a[i] % 2) != (i % 2)) {
      cout << -1 << '\n';
      return;
    }
  }
  auto find = [&](long long x) {
    for (long long i = 1; i <= n; i++) {
      if (a[i] == x) return i;
    }
    return x;
  };
  vector<long long> ans;
  auto doIt = [&](long long p) {
    assert(p & 1);
    ans.push_back(p);
    reverse(a.begin() + 1, a.begin() + p + 1);
  };
  for (long long r = n; r > 1; r -= 2) {
    long long p;
    p = find(r);
    doIt(p);
    p = find(r - 1) - 1;
    doIt(p);
    p = find(r - 1) + 1;
    doIt(p);
    p = 3;
    doIt(p);
    p = r;
    doIt(p);
  }
  assert(is_sorted(begin(a), end(a)));
  cout << (long long)ans.size() << '\n';
  for (long long &i : ans) cout << i << ' ';
  cout << '\n';
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long t_c = 1;
  cin >> t_c;
  while (t_c--) testCase();
}

```