#include <bits/stdc++.h>
#pragma GCC optimize(3)
using namespace std;
const long long N = 2199;
inline long long read() {
  long long s = 0, w = 1;
  register char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    s = (s << 3) + (s << 1) + (ch ^ 48), ch = getchar();
  return s * w;
}
long long n, a[N];
signed main() {
  for (register long long T = read(); T; T--) {
    n = read();
    for (register long long i = 1; i <= n; i++) a[i] = read();
    long long flg = 1;
    for (register long long i = 1; i <= n; i++) {
      if (abs(a[i] - i) & 1) {
        flg = 0;
        break;
      }
    }
    if (!flg) {
      puts("-1");
      continue;
    }
    vector<long long> g, w;
    for (register long long i = 1; i <= n; i++) w.emplace_back(a[i]);
    for (register long long i = n; i >= 3; i--, i--) {
      long long now = 0;
      for (register long long j = 1; j <= n; j++) {
        if (w[j - 1] == i) {
          now = j;
          break;
        }
      }
      g.emplace_back(now);
      reverse(w.begin(), w.begin() + now);
      long long pp = 0;
      for (register long long j = 1; j <= n; j++) {
        if (w[j - 1] == i - 1) {
          pp = j;
          break;
        }
      }
      g.emplace_back(pp - 1), reverse(w.begin(), w.begin() + pp - 1);
      g.emplace_back(pp + 1), reverse(w.begin(), w.begin() + pp + 1);
      g.emplace_back(3), reverse(w.begin(), w.begin() + 3);
      g.emplace_back(i), reverse(w.begin(), w.begin() + i);
    }
    printf("%lld\n", (long long)g.size());
    for (auto i : g) printf("%lld ", i);
    puts("");
  }
  return 0;
}

### Prompt

Your task is to create 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>
#pragma GCC optimize(3)
using namespace std;
const long long N = 2199;
inline long long read() {
  long long s = 0, w = 1;
  register char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    s = (s << 3) + (s << 1) + (ch ^ 48), ch = getchar();
  return s * w;
}
long long n, a[N];
signed main() {
  for (register long long T = read(); T; T--) {
    n = read();
    for (register long long i = 1; i <= n; i++) a[i] = read();
    long long flg = 1;
    for (register long long i = 1; i <= n; i++) {
      if (abs(a[i] - i) & 1) {
        flg = 0;
        break;
      }
    }
    if (!flg) {
      puts("-1");
      continue;
    }
    vector<long long> g, w;
    for (register long long i = 1; i <= n; i++) w.emplace_back(a[i]);
    for (register long long i = n; i >= 3; i--, i--) {
      long long now = 0;
      for (register long long j = 1; j <= n; j++) {
        if (w[j - 1] == i) {
          now = j;
          break;
        }
      }
      g.emplace_back(now);
      reverse(w.begin(), w.begin() + now);
      long long pp = 0;
      for (register long long j = 1; j <= n; j++) {
        if (w[j - 1] == i - 1) {
          pp = j;
          break;
        }
      }
      g.emplace_back(pp - 1), reverse(w.begin(), w.begin() + pp - 1);
      g.emplace_back(pp + 1), reverse(w.begin(), w.begin() + pp + 1);
      g.emplace_back(3), reverse(w.begin(), w.begin() + 3);
      g.emplace_back(i), reverse(w.begin(), w.begin() + i);
    }
    printf("%lld\n", (long long)g.size());
    for (auto i : g) printf("%lld ", i);
    puts("");
  }
  return 0;
}

```