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 Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << " : " << arg1 << std::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...);
}
template <class T>
ostream& operator<<(ostream& os, vector<T> V) {
  os << "[ ";
  for (auto v : V) os << v << " ";
  return os << "]";
}
template <class L, class R>
ostream& operator<<(ostream& os, pair<L, R> P) {
  return os << "(" << P.first << "," << P.second << ")";
}
template <typename T, typename U>
pair<T, U> operator+(const pair<T, U>& l, const std::pair<T, U>& r) {
  return {l.first + r.first, l.second + r.second};
}
const long long int mod = 1e9 + 7;
const int maxn = 100005;
struct DSU {
  vector<long long int> par, sz;
  DSU(long long int n) {
    par.resize(n);
    sz = vector<long long int>(n, 1);
    iota(par.begin(), par.end(), 0);
  }
  long long int find(long long int x) {
    if (x == par[x]) return x;
    return par[x] = find(par[x]);
  }
  void unite(long long int a, long long int b) {
    a = find(a);
    b = find(b);
    if (a == b) return;
    if (sz[a] < sz[b]) swap(a, b);
    par[b] = a;
    sz[a] += sz[b];
    return;
  }
};
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int m, n;
  cin >> m >> n;
  long long int a[m + 1], b[n + 1];
  DSU dsu(n + m + 2);
  for (long long int i = 1; i <= m; i++) {
    cin >> a[i];
  }
  for (long long int i = 1; i <= n; i++) {
    cin >> b[i];
  }
  long long int sum = 0;
  vector<tuple<long long int, long long int, long long int>> edges;
  for (long long int i = 1; i <= m; i++) {
    long long int k;
    cin >> k;
    while (k--) {
      long long int x;
      cin >> x;
      sum += a[i] + b[x];
      edges.push_back({a[i] + b[x], i, m + x});
    }
  }
  sort((edges).begin(), (edges).end());
  reverse((edges).begin(), (edges).end());
  for (long long int i = 0; i < edges.size(); i++) {
    long long int w, u, v;
    tie(w, u, v) = edges[i];
    if (dsu.find(u) == dsu.find(v))
      continue;
    else {
      sum -= w;
      dsu.unite(u, v);
    }
  }
  cout << sum << '\n';
}
