#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
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;
}
void write(long long x) {
  if (x < 0) putchar('-'), x = -x;
  if (x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}
const long long N = 1000010, mod = 1e9 + 7;
char s[N];
long long n, a[N], A, B, xianA, xianB;
vector<long long> ans;
long long find(long long x) {
  long long ans = 0;
  for (long long i = (long long)(1); i <= (long long)(n); ++i)
    if (a[i] == x) ans = i;
  return ans;
}
int main() {
  for (long long T = read(); T--;) {
    n = read();
    bool fl = 1;
    ans.clear();
    for (long long i = (long long)(1); i <= (long long)(n); ++i) {
      a[i] = read();
      fl &= (a[i] & 1) == (i & 1);
    }
    if (!fl)
      puts("-1");
    else {
      for (long long i = n; i >= 3; i -= 2) {
        long long p;
        p = find(i);
        if (p > 0 && p != 1) ans.push_back(p), reverse(a + 1, a + p + 1);
        p = find(i - 1) - 1;
        if (p > 0 && p != 1) ans.push_back(p), reverse(a + 1, a + p + 1);
        p = find(i - 1) + 1;
        if (p > 0 && p != 1) ans.push_back(p), reverse(a + 1, a + p + 1);
        p = find(i);
        if (p > 0 && p != 1) ans.push_back(p), reverse(a + 1, a + p + 1);
        ans.push_back(i);
        reverse(a + 1, a + i + 1);
      }
      cout << ans.size() << endl;
      for (long long i = (long long)(0); i < (long long)(ans.size()); ++i)
        write(ans[i]), putchar(' ');
      puts("");
    }
  }
}

### Prompt

Develop a solution in cpp to the problem described below:

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("Ofast")
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;
}
void write(long long x) {
  if (x < 0) putchar('-'), x = -x;
  if (x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}
const long long N = 1000010, mod = 1e9 + 7;
char s[N];
long long n, a[N], A, B, xianA, xianB;
vector<long long> ans;
long long find(long long x) {
  long long ans = 0;
  for (long long i = (long long)(1); i <= (long long)(n); ++i)
    if (a[i] == x) ans = i;
  return ans;
}
int main() {
  for (long long T = read(); T--;) {
    n = read();
    bool fl = 1;
    ans.clear();
    for (long long i = (long long)(1); i <= (long long)(n); ++i) {
      a[i] = read();
      fl &= (a[i] & 1) == (i & 1);
    }
    if (!fl)
      puts("-1");
    else {
      for (long long i = n; i >= 3; i -= 2) {
        long long p;
        p = find(i);
        if (p > 0 && p != 1) ans.push_back(p), reverse(a + 1, a + p + 1);
        p = find(i - 1) - 1;
        if (p > 0 && p != 1) ans.push_back(p), reverse(a + 1, a + p + 1);
        p = find(i - 1) + 1;
        if (p > 0 && p != 1) ans.push_back(p), reverse(a + 1, a + p + 1);
        p = find(i);
        if (p > 0 && p != 1) ans.push_back(p), reverse(a + 1, a + p + 1);
        ans.push_back(i);
        reverse(a + 1, a + i + 1);
      }
      cout << ans.size() << endl;
      for (long long i = (long long)(0); i < (long long)(ans.size()); ++i)
        write(ans[i]), putchar(' ');
      puts("");
    }
  }
}

```