#include <bits/stdc++.h>
using namespace std;
const int MAXN = 2105 * 5;
int n;
int a[MAXN];
int ans[MAXN], anstot;
long long Read() {
  long long x = 0, f = 1;
  char c = getchar();
  while (c > '9' || c < '0') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = (x * 10) + (c ^ 48);
    c = getchar();
  }
  return x * f;
}
template <typename T>
void Put1(T x) {
  if (x > 9) Put1(x / 10);
  putchar(x % 10 ^ 48);
}
template <typename T>
void Put(T x, char c = -1) {
  if (x < 0) putchar('-'), x = -x;
  Put1(x);
  if (c >= 0) putchar(c);
}
template <typename T>
T Max(T x, T y) {
  return x > y ? x : y;
}
template <typename T>
T Min(T x, T y) {
  return x < y ? x : y;
}
template <typename T>
T Abs(T x) {
  return x < 0 ? -x : x;
}
void solve(int x) {
  if (x == 1) return;
  for (int i = 1; i <= x / 2; ++i) swap(a[i], a[x - i + 1]);
  ans[++anstot] = x;
}
int main() {
  for (int T = Read(); T; --T) {
    n = Read();
    bool ls = 0;
    for (int i = 1; i <= n; ++i) {
      a[i] = Read();
      if ((a[i] & 1) ^ (i & 1)) ls = 1;
    }
    if (ls) {
      Put(-1, '\n');
      continue;
    }
    for (int i = 1; i <= n; ++i)
      if (a[i] ^ i) ls = 1;
    if (!ls) {
      Put(0, '\n');
      continue;
    }
    anstot = 0;
    for (int i = 1; i < n; i += 2) {
      int ID1 = 0, ID2 = 0;
      for (int j = 1; j <= n; ++j)
        if (a[j] == i)
          ID1 = j;
        else if (a[j] == i + 1)
          ID2 = j;
      if (ID1 > ID2) ID2 = ID1 - ID2 + 1;
      solve(ID1);
      solve(ID2 - 1);
      ID1 = ID2 - 1;
      int val = ID2 + 1;
      ID1 = val - ID1 + 1;
      ID2 = val - ID2 + 1;
      solve(val);
      solve(ID1);
      ID1 = 1;
      ID2 = 2;
      solve(n - i + 1);
    }
    solve(n);
    Put(anstot, '\n');
    for (int i = 1; i <= anstot; ++i) Put(ans[i], ' ');
    putchar('\n');
  }
  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;
const int MAXN = 2105 * 5;
int n;
int a[MAXN];
int ans[MAXN], anstot;
long long Read() {
  long long x = 0, f = 1;
  char c = getchar();
  while (c > '9' || c < '0') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = (x * 10) + (c ^ 48);
    c = getchar();
  }
  return x * f;
}
template <typename T>
void Put1(T x) {
  if (x > 9) Put1(x / 10);
  putchar(x % 10 ^ 48);
}
template <typename T>
void Put(T x, char c = -1) {
  if (x < 0) putchar('-'), x = -x;
  Put1(x);
  if (c >= 0) putchar(c);
}
template <typename T>
T Max(T x, T y) {
  return x > y ? x : y;
}
template <typename T>
T Min(T x, T y) {
  return x < y ? x : y;
}
template <typename T>
T Abs(T x) {
  return x < 0 ? -x : x;
}
void solve(int x) {
  if (x == 1) return;
  for (int i = 1; i <= x / 2; ++i) swap(a[i], a[x - i + 1]);
  ans[++anstot] = x;
}
int main() {
  for (int T = Read(); T; --T) {
    n = Read();
    bool ls = 0;
    for (int i = 1; i <= n; ++i) {
      a[i] = Read();
      if ((a[i] & 1) ^ (i & 1)) ls = 1;
    }
    if (ls) {
      Put(-1, '\n');
      continue;
    }
    for (int i = 1; i <= n; ++i)
      if (a[i] ^ i) ls = 1;
    if (!ls) {
      Put(0, '\n');
      continue;
    }
    anstot = 0;
    for (int i = 1; i < n; i += 2) {
      int ID1 = 0, ID2 = 0;
      for (int j = 1; j <= n; ++j)
        if (a[j] == i)
          ID1 = j;
        else if (a[j] == i + 1)
          ID2 = j;
      if (ID1 > ID2) ID2 = ID1 - ID2 + 1;
      solve(ID1);
      solve(ID2 - 1);
      ID1 = ID2 - 1;
      int val = ID2 + 1;
      ID1 = val - ID1 + 1;
      ID2 = val - ID2 + 1;
      solve(val);
      solve(ID1);
      ID1 = 1;
      ID2 = 2;
      solve(n - i + 1);
    }
    solve(n);
    Put(anstot, '\n');
    for (int i = 1; i <= anstot; ++i) Put(ans[i], ' ');
    putchar('\n');
  }
  return 0;
}

```