#include <bits/stdc++.h>
using namespace std;
void cppio() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
}
const int maxn = int(2e5) + 10;
int n;
int a[maxn];
int ans[maxn], an;
void Swap(int r) {
  int l = 1;
  if (l == r) return;
  ans[an++] = r;
  reverse(a + l, a + r + 1);
}
void Work() {
  cin >> n;
  for (int i = 1; i <= (n); ++i) cin >> a[i];
  for (int i = 1; i <= (n); ++i)
    if ((i & 1) != (a[i] & 1)) {
      cout << "-1\n";
      return;
    }
  an = 0;
  for (int z = n; 1 < z; z -= 2) {
    int zp = find(a + 1, a + z + 1, z) - a;
    int z2p = find(a + 2, a + z, z - 1) - a;
    if (zp < z2p) {
      Swap(zp);
      Swap(z2p - 1);
      Swap(z2p + 1);
      Swap(3);
      Swap(z);
    } else {
      Swap(zp);
      z2p = 1 + zp - z2p;
      Swap(z2p - 1);
      Swap(z2p + 1);
      Swap(3);
      Swap(z);
    }
  }
  cout << an << '\n';
  for (int i = 0; i < (an); ++i) cout << ans[i] << " \n"[i + 1 == an];
}
int main() {
  cppio();
  int tc;
  cin >> tc;
  for (int tci = 1; tci <= (tc); ++tci) {
    Work();
  }
  return 0;
}

### 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;
void cppio() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
}
const int maxn = int(2e5) + 10;
int n;
int a[maxn];
int ans[maxn], an;
void Swap(int r) {
  int l = 1;
  if (l == r) return;
  ans[an++] = r;
  reverse(a + l, a + r + 1);
}
void Work() {
  cin >> n;
  for (int i = 1; i <= (n); ++i) cin >> a[i];
  for (int i = 1; i <= (n); ++i)
    if ((i & 1) != (a[i] & 1)) {
      cout << "-1\n";
      return;
    }
  an = 0;
  for (int z = n; 1 < z; z -= 2) {
    int zp = find(a + 1, a + z + 1, z) - a;
    int z2p = find(a + 2, a + z, z - 1) - a;
    if (zp < z2p) {
      Swap(zp);
      Swap(z2p - 1);
      Swap(z2p + 1);
      Swap(3);
      Swap(z);
    } else {
      Swap(zp);
      z2p = 1 + zp - z2p;
      Swap(z2p - 1);
      Swap(z2p + 1);
      Swap(3);
      Swap(z);
    }
  }
  cout << an << '\n';
  for (int i = 0; i < (an); ++i) cout << ans[i] << " \n"[i + 1 == an];
}
int main() {
  cppio();
  int tc;
  cin >> tc;
  for (int tci = 1; tci <= (tc); ++tci) {
    Work();
  }
  return 0;
}

```