#include <bits/stdc++.h>
using namespace std;
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int a[3333], cnt = 0, p[3333];
vector<int> ans;
bool odd(int x) { return (x & 1) ? 1 : 0; }
void upd(int x) {
  if (x > 1) ans.push_back(x);
  for (register int i = 1; i <= x; ++i) p[a[i]] = i;
}
void work(int k) {
  reverse(a + 1, a + p[k] + 1);
  upd(p[k]);
  reverse(a + 1, a + p[k - 1] - 1 + 1);
  upd(p[k - 1] - 1);
  reverse(a + 1, a + p[k - 1] + 1 + 1);
  upd(p[k - 1] + 1);
  reverse(a + 1, a + p[k] + 1);
  upd(p[k]);
  reverse(a + 1, a + k + 1);
  upd(k);
  if (k > 1) work(k - 2);
}
int main() {
  int t = read();
  while (t--) {
    int n = read();
    bool flag = 1;
    for (register int i = 1; i <= n; ++i) {
      a[i] = read(), p[a[i]] = i;
      if (odd(i) ^ odd(a[i])) flag = 0;
    }
    if (!flag) {
      puts("-1");
      continue;
    }
    cnt = 0, ans.clear();
    work(n);
    printf("%d\n", (int)ans.size());
    for (int x : ans) printf("%d ", x);
    puts("");
  }
  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>
using namespace std;
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int a[3333], cnt = 0, p[3333];
vector<int> ans;
bool odd(int x) { return (x & 1) ? 1 : 0; }
void upd(int x) {
  if (x > 1) ans.push_back(x);
  for (register int i = 1; i <= x; ++i) p[a[i]] = i;
}
void work(int k) {
  reverse(a + 1, a + p[k] + 1);
  upd(p[k]);
  reverse(a + 1, a + p[k - 1] - 1 + 1);
  upd(p[k - 1] - 1);
  reverse(a + 1, a + p[k - 1] + 1 + 1);
  upd(p[k - 1] + 1);
  reverse(a + 1, a + p[k] + 1);
  upd(p[k]);
  reverse(a + 1, a + k + 1);
  upd(k);
  if (k > 1) work(k - 2);
}
int main() {
  int t = read();
  while (t--) {
    int n = read();
    bool flag = 1;
    for (register int i = 1; i <= n; ++i) {
      a[i] = read(), p[a[i]] = i;
      if (odd(i) ^ odd(a[i])) flag = 0;
    }
    if (!flag) {
      puts("-1");
      continue;
    }
    cnt = 0, ans.clear();
    work(n);
    printf("%d\n", (int)ans.size());
    for (int x : ans) printf("%d ", x);
    puts("");
  }
  return 0;
}

```