You are given m sets of integers A_1, A_2, …, A_m; elements of these sets are integers between 1 and n, inclusive.

There are two arrays of positive integers a_1, a_2, …, a_m and b_1, b_2, …, b_n. 

In one operation you can delete an element j from the set A_i and pay a_i + b_j coins for that.

You can make several (maybe none) operations (some sets can become empty).

After that, you will make an edge-colored undirected graph consisting of n vertices. For each set A_i you will add an edge (x, y) with color i for all x, y ∈ A_i and x < y. Some pairs of vertices can be connected with more than one edge, but such edges have different colors.

You call a cycle i_1 → e_1 → i_2 → e_2 → … → i_k → e_k → i_1 (e_j is some edge connecting vertices i_j and i_{j+1} in this graph) rainbow if all edges on it have different colors.

Find the minimum number of coins you should pay to get a graph without rainbow cycles.

Input

The first line contains two integers m and n (1 ≤ m, n ≤ 10^5), the number of sets and the number of vertices in the graph.

The second line contains m integers a_1, a_2, …, a_m (1 ≤ a_i ≤ 10^9).

The third line contains n integers b_1, b_2, …, b_n (1 ≤ b_i ≤ 10^9).

In the each of the next of m lines there are descriptions of sets. In the i-th line the first integer s_i (1 ≤ s_i ≤ n) is equal to the size of A_i. Then s_i integers follow: the elements of the set A_i. These integers are from 1 to n and distinct.

It is guaranteed that the sum of s_i for all 1 ≤ i ≤ m does not exceed 2 ⋅ 10^5.

Output

Print one integer: the minimum number of coins you should pay for operations to avoid rainbow cycles in the obtained graph.

Examples

Input


3 2
1 2 3
4 5
2 1 2
2 1 2
2 1 2


Output


11


Input


7 8
3 6 7 9 10 7 239
8 1 9 7 10 2 6 239
3 2 1 3
2 4 1
3 1 3 7
2 4 3
5 3 4 5 6 7
2 5 7
1 8


Output


66

Note

In the first test, you can make such operations:

  * Delete element 1 from set 1. You should pay a_1 + b_1 = 5 coins for that. 
  * Delete element 1 from set 2. You should pay a_2 + b_1 = 6 coins for that. 



You pay 11 coins in total. After these operations, the first and the second sets will be equal to \{2\} and the third set will be equal to \{1, 2\}.

So, the graph will consist of one edge (1, 2) of color 3.

In the second test, you can make such operations:

  * Delete element 1 from set 1. You should pay a_1 + b_1 = 11 coins for that. 
  * Delete element 4 from set 2. You should pay a_2 + b_4 = 13 coins for that. 
  * Delete element 7 from set 3. You should pay a_3 + b_7 = 13 coins for that. 
  * Delete element 4 from set 4. You should pay a_4 + b_4 = 16 coins for that. 
  * Delete element 7 from set 6. You should pay a_6 + b_7 = 13 coins for that. 



You pay 66 coins in total.

After these operations, the sets will be:

  * \{2, 3\}; 
  * \{1\}; 
  * \{1, 3\}; 
  * \{3\}; 
  * \{3, 4, 5, 6, 7\}; 
  * \{5\}; 
  * \{8\}. 



We will get the graph:

<image>

There are no rainbow cycles in it.
#include <bits/stdc++.h>
using namespace std;
void debug_out() { cout << '\n'; }
template <typename T, typename... R>
void debug_out(const T &f, const R &...r) {
  cout << f << " ";
  debug_out(r...);
}
template <typename A, typename B>
ostream &operator<<(ostream &out, const pair<A, B> &a) {
  out << "(" << a.first << "," << a.second << ")";
  return out;
}
template <typename Tuple, size_t N>
struct tuple_print {
  static void print(const Tuple &t, std::ostream &os) {
    tuple_print<Tuple, N - 1>::print(t, os);
    os << ", " << std::get<N - 1>(t);
  }
};
template <typename Tuple>
struct tuple_print<Tuple, 1> {
  static void print(const Tuple &t, std::ostream &os) {
    os << "(" << std::get<0>(t);
  }
};
template <typename... Args>
ostream &operator<<(ostream &os, const tuple<Args...> &t) {
  tuple_print<decltype(t), sizeof...(Args)>::print(t, os);
  return os << ")";
}
template <typename T>
ostream &operator<<(ostream &out, const vector<T> &a) {
  out << "[";
  bool first = true;
  for (auto &v : a) {
    out << (first ? "" : ", ");
    out << v;
    first = 0;
  }
  out << "]";
  return out;
}
template <typename T, class Cmp>
ostream &operator<<(ostream &out, const set<T, Cmp> &a) {
  out << "{";
  bool first = true;
  for (auto &v : a) {
    out << (first ? "" : ", ");
    out << v;
    first = 0;
  }
  out << "}";
  return out;
}
template <typename U, typename T, class Cmp>
ostream &operator<<(ostream &out, const map<U, T, Cmp> &a) {
  out << "{";
  bool first = true;
  for (auto &p : a) {
    out << (first ? "" : ", ");
    out << p.first << ":" << p.second;
    first = 0;
  }
  out << "}";
  return out;
}
template <typename Arg1>
void __f(const char *name, Arg1 &&arg1) {
  cerr << name << ": " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char *names, Arg1 &&arg1, Args &&...args) {
  const char *comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << ": " << arg1 << " |";
  __f(comma + 1, args...);
}
const long long mod = 1e9 + 7;
const int N = 1e6 + 100;
int T;
int m, n;
long long a[N], b[N];
int fa[N], s, x;
tuple<int, int, long long> edge[N];
int getfa(int xx) {
  if (xx == fa[xx]) return xx;
  return fa[xx] = getfa(fa[xx]);
}
int main() {
  ios::sync_with_stdio(false);
  cin >> m >> n;
  for (int i = 1; i <= m; i++) cin >> a[i];
  for (int i = 1; i <= n; i++) cin >> b[i];
  int all = 0;
  long long sum = 0;
  for (int i = 1; i <= m; i++) {
    cin >> s;
    for (int j = 1; j <= s; j++) {
      cin >> x;
      edge[++all] = make_tuple(n + i, x, a[i] + b[x]);
      sum += a[i] + b[x];
    }
  }
  sort(edge + 1, edge + all + 1,
       [&](tuple<int, int, long long> aa, tuple<int, int, long long> bb) {
         return get<2>(aa) > get<2>(bb);
       });
  for (int i = 1; i <= n + m; i++) fa[i] = i;
  long long ans = 0;
  for (int i = 1; i <= all; i++) {
    int yy = getfa(get<1>(edge[i]));
    int xx = getfa(get<0>(edge[i]));
    if (xx == yy) continue;
    fa[xx] = yy;
    ans += get<2>(edge[i]);
  }
  cout << sum - ans;
  return 0;
}
