#include <bits/stdc++.h>
using namespace std;
struct chash {
  static inline uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  inline size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
  inline size_t operator()(const pair<uint64_t, uint64_t> &v) const {
    chash int_hasher;
    return int_hasher(v.first) * 31 + int_hasher(v.second);
  }
};
const char nl = '\n';
const long long inf = 1e9 + 10;
const long long inf2 = 1e18 + 99LL;
const long double inf3 = 1e17;
const long long mod = 1e9 + 7, mod2 = 998244353;
const long double eps = 1e-9;
const bool local = false;
const int logn = 21, maxn = 200001, maxm = 1, maxn2 = 3;
int a[maxn];
void f(int x) {
  for (int i = 0; i < x / 2; i++) {
    swap(a[i], a[x - i - 1]);
  }
}
void solve() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (int i = 0; i < n; i++) {
    if (a[i] % 2 != (i + 1) % 2) {
      cout << -1 << nl;
      return;
    }
  }
  vector<int> ans;
  for (int i = n; i > 1; i -= 2) {
    if (a[i - 1] == i && a[i - 2] == i - 1) continue;
    int i1 = 0, i2 = 0;
    for (int j = 0; j < n; j++) {
      if (a[j] == i)
        i1 = j;
      else if (a[j] == i - 1)
        i2 = j;
    }
    if (i1 + 1 > 1) {
      ans.push_back(i1 + 1);
      f(i1 + 1);
      if (i2 < i1) i2 = i1 - i2;
      i1 = 0;
    }
    if (i2 > 1) {
      ans.push_back(i2);
      f(i2);
    }
    i1 = i2 - 1;
    f(i);
    ans.push_back(i);
    i2 = i - i2 - 1;
    f(i2 + 2);
    ans.push_back(i2 + 2);
    f(i);
    ans.push_back(i);
  }
  cout << (int)((ans).size()) << nl;
  for (int x : ans) cout << x << ' ';
  cout << nl;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  int T;
  cin >> T;
  while (T--) solve();
}

### 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>
using namespace std;
struct chash {
  static inline uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  inline size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
  inline size_t operator()(const pair<uint64_t, uint64_t> &v) const {
    chash int_hasher;
    return int_hasher(v.first) * 31 + int_hasher(v.second);
  }
};
const char nl = '\n';
const long long inf = 1e9 + 10;
const long long inf2 = 1e18 + 99LL;
const long double inf3 = 1e17;
const long long mod = 1e9 + 7, mod2 = 998244353;
const long double eps = 1e-9;
const bool local = false;
const int logn = 21, maxn = 200001, maxm = 1, maxn2 = 3;
int a[maxn];
void f(int x) {
  for (int i = 0; i < x / 2; i++) {
    swap(a[i], a[x - i - 1]);
  }
}
void solve() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (int i = 0; i < n; i++) {
    if (a[i] % 2 != (i + 1) % 2) {
      cout << -1 << nl;
      return;
    }
  }
  vector<int> ans;
  for (int i = n; i > 1; i -= 2) {
    if (a[i - 1] == i && a[i - 2] == i - 1) continue;
    int i1 = 0, i2 = 0;
    for (int j = 0; j < n; j++) {
      if (a[j] == i)
        i1 = j;
      else if (a[j] == i - 1)
        i2 = j;
    }
    if (i1 + 1 > 1) {
      ans.push_back(i1 + 1);
      f(i1 + 1);
      if (i2 < i1) i2 = i1 - i2;
      i1 = 0;
    }
    if (i2 > 1) {
      ans.push_back(i2);
      f(i2);
    }
    i1 = i2 - 1;
    f(i);
    ans.push_back(i);
    i2 = i - i2 - 1;
    f(i2 + 2);
    ans.push_back(i2 + 2);
    f(i);
    ans.push_back(i);
  }
  cout << (int)((ans).size()) << nl;
  for (int x : ans) cout << x << ' ';
  cout << nl;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  int T;
  cin >> T;
  while (T--) solve();
}

```