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;
template <typename T>
class graph {
 public:
  struct edge {
    int from;
    int to;
    T cost;
  };
  vector<edge> edges;
  vector<vector<int>> g;
  int n;
  graph(int _n) : n(_n) { g.resize(n); }
  virtual int add(int from, int to, T cost) = 0;
};
template <typename T>
class undigraph : public graph<T> {
 public:
  using graph<T>::edges;
  using graph<T>::g;
  using graph<T>::n;
  undigraph(int _n) : graph<T>(_n) {}
  int add(int from, int to, T cost = 1) {
    assert(0 <= from && from < n && 0 <= to && to < n);
    int id = (int)edges.size();
    g[from].push_back(id);
    g[to].push_back(id);
    edges.push_back({from, to, cost});
    return id;
  }
};
class dsu {
 public:
  vector<int> p;
  int n;
  dsu(int _n) : n(_n) {
    p.resize(n);
    iota(p.begin(), p.end(), 0);
  }
  inline int get(int x) { return (x == p[x] ? x : (p[x] = get(p[x]))); }
  inline bool unite(int x, int y) {
    x = get(x);
    y = get(y);
    if (x != y) {
      p[x] = y;
      return true;
    }
    return false;
  }
};
template <typename T>
vector<int> find_mst(const undigraph<T> &g, T &ans) {
  vector<int> order(g.edges.size());
  iota(order.begin(), order.end(), 0);
  sort(order.begin(), order.end(),
       [&g](int a, int b) { return g.edges[a].cost > g.edges[b].cost; });
  dsu d(g.n);
  vector<int> ans_list;
  ans = 0;
  for (int id : order) {
    auto &e = g.edges[id];
    if (d.get(e.from) != d.get(e.to)) {
      d.unite(e.from, e.to);
      ans_list.push_back(id);
      ans += e.cost;
    }
  }
  return ans_list;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int m, n;
  cin >> m >> n;
  vector<int> a(m);
  for (int i = 0; i < m; i++) {
    cin >> a[i];
  }
  vector<int> b(n);
  for (int i = 0; i < n; i++) {
    cin >> b[i];
  }
  undigraph<long long> g(m + n);
  long long ans = 0;
  for (int i = 0; i < m; i++) {
    int foo;
    cin >> foo;
    while (foo--) {
      int j;
      cin >> j;
      --j;
      g.add(i, m + j, a[i] + b[j]);
      ans += a[i] + b[j];
    }
  }
  long long sub = 0;
  find_mst(g, sub);
  cout << ans - sub << '\n';
  return 0;
}
