#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma GCC target("sse4")
#pragma GCC optimize("unroll-loops")
using namespace std;
template <class T>
inline void amin(T &x, const T &y) {
  if (y < x) x = y;
}
template <class T>
inline void amax(T &x, const T &y) {
  if (x < y) x = y;
}
int N;
int A[2222];
vector<int> ans;
void ADD(int x) {
  assert(x % 2 == 1);
  ans.push_back(x);
  reverse(A, A + x);
}
void MAIN() {
  scanf("%d", &N);
  for (int i = 0, i_len = (N); i < i_len; ++i) scanf("%d", A + i);
  bool yes = true;
  for (int i = 0, i_len = (N); i < i_len; ++i)
    if (A[i] % 2 == i % 2) yes = false;
  if (yes) {
    ans.clear();
    int len = N;
    while (len >= 3) {
      if (A[len - 1] == len && A[len - 2] == len - 1) {
        len -= 2;
      } else if (A[0] == len && A[1] == len - 1) {
        ADD(len);
      } else {
        int a = -1, b = -1;
        for (int i = 0, i_len = (len); i < i_len; ++i) {
          if (A[i] == len) a = i;
          if (A[i] == len - 1) b = i;
        }
        if (b + 1 == a) {
          ADD(a + 1);
        } else if (a + 1 == b) {
          ADD(b + 2);
        } else if (a == 0) {
          ADD(b);
        } else {
          ADD(a + 1);
        }
      }
    }
    printf("%d\n", (int)ans.size());
    do {
      const auto end_rp = (ans.end());
      auto it_rp = (ans.begin());
      for (bool sp_rp = 0; it_rp != end_rp; ++it_rp) {
        if (sp_rp)
          putchar(' ');
        else
          sp_rp = true;
        printf("%d", *it_rp);
      }
      putchar('\n');
    } while (0);
  } else {
    puts("-1");
  }
}
int main() {
  int TC = 1;
  scanf("%d", &TC);
  for (int tc = 0, tc_len = (TC); tc < tc_len; ++tc) MAIN();
  return 0;
}

### Prompt

Generate 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("O3")
#pragma GCC target("sse4")
#pragma GCC optimize("unroll-loops")
using namespace std;
template <class T>
inline void amin(T &x, const T &y) {
  if (y < x) x = y;
}
template <class T>
inline void amax(T &x, const T &y) {
  if (x < y) x = y;
}
int N;
int A[2222];
vector<int> ans;
void ADD(int x) {
  assert(x % 2 == 1);
  ans.push_back(x);
  reverse(A, A + x);
}
void MAIN() {
  scanf("%d", &N);
  for (int i = 0, i_len = (N); i < i_len; ++i) scanf("%d", A + i);
  bool yes = true;
  for (int i = 0, i_len = (N); i < i_len; ++i)
    if (A[i] % 2 == i % 2) yes = false;
  if (yes) {
    ans.clear();
    int len = N;
    while (len >= 3) {
      if (A[len - 1] == len && A[len - 2] == len - 1) {
        len -= 2;
      } else if (A[0] == len && A[1] == len - 1) {
        ADD(len);
      } else {
        int a = -1, b = -1;
        for (int i = 0, i_len = (len); i < i_len; ++i) {
          if (A[i] == len) a = i;
          if (A[i] == len - 1) b = i;
        }
        if (b + 1 == a) {
          ADD(a + 1);
        } else if (a + 1 == b) {
          ADD(b + 2);
        } else if (a == 0) {
          ADD(b);
        } else {
          ADD(a + 1);
        }
      }
    }
    printf("%d\n", (int)ans.size());
    do {
      const auto end_rp = (ans.end());
      auto it_rp = (ans.begin());
      for (bool sp_rp = 0; it_rp != end_rp; ++it_rp) {
        if (sp_rp)
          putchar(' ');
        else
          sp_rp = true;
        printf("%d", *it_rp);
      }
      putchar('\n');
    } while (0);
  } else {
    puts("-1");
  }
}
int main() {
  int TC = 1;
  scanf("%d", &TC);
  for (int tc = 0, tc_len = (TC); tc < tc_len; ++tc) MAIN();
  return 0;
}

```