#include <bits/stdc++.h>
using namespace std;
void abc() { cerr << endl; }
template <typename T, typename... U>
void abc(T a, U... b) {
  cerr << a << ' ', abc(b...);
}
template <typename T>
void printv(T l, T r) {
  while (l != r) cout << *l << " \n"[++l == r];
}
template <typename A, typename B>
istream& operator>>(istream& o, pair<A, B>& a) {
  return o >> a.first >> a.second;
}
template <typename A, typename B>
ostream& operator<<(ostream& o, pair<A, B> a) {
  return o << '(' << a.first << ", " << a.second << ')';
}
template <typename T>
ostream& operator<<(ostream& o, vector<T> a) {
  bool is = false;
  for (T i : a) {
    o << (is ? ' ' : '{'), is = true, o << i;
  }
  return o << '}';
}
template <typename T>
struct vv : vector<vector<T>> {
  vv(int n, int m, T v) : vector<vector<T>>(n, vector<T>(m, v)) {}
  vv() {}
};
template <typename T>
struct vvv : vector<vv<T>> {
  vvv(int n, int m, int k, T v) : vector<vv<T>>(n, vv<T>(m, k, v)) {}
  vvv() {}
};
const int N = 200000, logN = 18, K = 1000;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<int> p(n);
    for (int i = 0; i < n; ++i) cin >> p[i];
    vector<int> ans;
    auto add = [&](int x) {
      ans.push_back(x);
      reverse(p.begin(), p.begin() + x);
    };
    bool is = true;
    for (int now = n; now > 1; now -= 2) {
      int f = -1, s = -1;
      for (int i = 0; i < now; ++i) {
        if (p[i] == now - 1) f = i;
        if (p[i] == now) s = i;
      }
      if (f & 1 && s & 1 ^ 1) {
        add(s + 1);
        for (int i = 0; i < now; ++i) {
          if (p[i] == now - 1) f = i;
          if (p[i] == now) s = i;
        }
        if (s + 1 != f) {
          add(f);
          for (int i = 0; i < now; ++i) {
            if (p[i] == now - 1) f = i;
            if (p[i] == now) s = i;
          }
        }
        add(now);
        for (int i = 0; i < now; ++i) {
          if (p[i] == now - 1) f = i;
          if (p[i] == now) s = i;
        }
        add(s + 1);
        add(now);
      } else {
        is = false;
        break;
      }
    }
    if (is) {
      void(0);
      cout << ans.size() << endl;
      printv(ans.begin(), ans.end());
    } else {
      cout << -1 << endl;
    }
  }
}

### 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;
void abc() { cerr << endl; }
template <typename T, typename... U>
void abc(T a, U... b) {
  cerr << a << ' ', abc(b...);
}
template <typename T>
void printv(T l, T r) {
  while (l != r) cout << *l << " \n"[++l == r];
}
template <typename A, typename B>
istream& operator>>(istream& o, pair<A, B>& a) {
  return o >> a.first >> a.second;
}
template <typename A, typename B>
ostream& operator<<(ostream& o, pair<A, B> a) {
  return o << '(' << a.first << ", " << a.second << ')';
}
template <typename T>
ostream& operator<<(ostream& o, vector<T> a) {
  bool is = false;
  for (T i : a) {
    o << (is ? ' ' : '{'), is = true, o << i;
  }
  return o << '}';
}
template <typename T>
struct vv : vector<vector<T>> {
  vv(int n, int m, T v) : vector<vector<T>>(n, vector<T>(m, v)) {}
  vv() {}
};
template <typename T>
struct vvv : vector<vv<T>> {
  vvv(int n, int m, int k, T v) : vector<vv<T>>(n, vv<T>(m, k, v)) {}
  vvv() {}
};
const int N = 200000, logN = 18, K = 1000;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<int> p(n);
    for (int i = 0; i < n; ++i) cin >> p[i];
    vector<int> ans;
    auto add = [&](int x) {
      ans.push_back(x);
      reverse(p.begin(), p.begin() + x);
    };
    bool is = true;
    for (int now = n; now > 1; now -= 2) {
      int f = -1, s = -1;
      for (int i = 0; i < now; ++i) {
        if (p[i] == now - 1) f = i;
        if (p[i] == now) s = i;
      }
      if (f & 1 && s & 1 ^ 1) {
        add(s + 1);
        for (int i = 0; i < now; ++i) {
          if (p[i] == now - 1) f = i;
          if (p[i] == now) s = i;
        }
        if (s + 1 != f) {
          add(f);
          for (int i = 0; i < now; ++i) {
            if (p[i] == now - 1) f = i;
            if (p[i] == now) s = i;
          }
        }
        add(now);
        for (int i = 0; i < now; ++i) {
          if (p[i] == now - 1) f = i;
          if (p[i] == now) s = i;
        }
        add(s + 1);
        add(now);
      } else {
        is = false;
        break;
      }
    }
    if (is) {
      void(0);
      cout << ans.size() << endl;
      printv(ans.begin(), ans.end());
    } else {
      cout << -1 << endl;
    }
  }
}

```