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>
#pragma GCC target("avx2,avx")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
using namespace std;
using ll = long long;
using ull = unsigned long long;
using pii = pair<int, int>;
using pll = pair<long long, long long>;
constexpr char ln = '\n';
template <class T1, class T2>
inline bool chmax(T1 &a, T2 b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <class T1, class T2>
inline bool chmin(T1 &a, T2 b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
inline int topbit(int x) { return x == 0 ? -1 : 31 - __builtin_clz(x); }
inline int topbit(long long x) { return x == 0 ? -1 : 63 - __builtin_clzll(x); }
inline int botbit(int x) { return x == 0 ? 32 : __builtin_ctz(x); }
inline int botbit(long long x) { return x == 0 ? 64 : __builtin_ctzll(x); }
inline int popcount(int x) { return __builtin_popcount(x); }
inline int popcount(long long x) { return __builtin_popcountll(x); }
inline int kthbit(long long x, int k) { return (x >> k) & 1; }
inline void print() { cout << "\n"; }
template <class T>
inline void print(const vector<T> &v) {
  for (auto itr = v.begin(); itr != v.end(); ++itr) cout << *itr << " ";
  print();
}
template <class T, class... Args>
inline void print(const T &x, const Args &...args) {
  cout << x << " ";
  print(args...);
}
inline void debug() {}
template <class T>
inline void debug(const vector<T> &v) {}
template <class T, class... Args>
inline void debug(const T &x, const Args &...args) {}
struct Fast_ios {
  Fast_ios() {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(20);
  };
} fast_ios;
struct UnionFind {
  int N;
  vector<int> node;
  UnionFind() {}
  UnionFind(int N) : N(N), node(N, -1) {}
  void init(int x) {
    node.assign(x, -1);
    N = x;
  }
  bool merge(int x, int y) {
    x = root(x);
    y = root(y);
    if (x == y) return false;
    if (node[x] > node[y]) swap(x, y);
    node[x] += node[y];
    node[y] = x;
    N--;
    return true;
  }
  bool same(int x, int y) { return root(x) == root(y); }
  int root(int x) {
    if (node[x] < 0) return x;
    return node[x] = root(node[x]);
  }
  int size(int x) { return -node[root(x)]; }
  int count() const { return N; }
};
template <typename T>
struct edge {
  int src, to;
  T cost;
  int id;
  edge() {}
  edge(int to, T cost) : src(-1), to(to), cost(cost), id(-1) {}
  edge(int src, int to, T cost) : src(src), to(to), cost(cost), id(-1) {}
  edge(int src, int to, T cost, int id)
      : src(src), to(to), cost(cost), id(id) {}
  bool operator<(const edge &e) const { return cost < e.cost; }
};
int main() {
  int M, N;
  cin >> M >> N;
  vector<ll> A(M), B(N);
  for (int i = 0; i < (M); ++i) cin >> A[i];
  for (int i = 0; i < (N); ++i) cin >> B[i];
  vector<edge<ll>> E;
  ll ans = 0;
  for (int i = 0; i < (M); ++i) {
    int cnt;
    cin >> cnt;
    for (int j = 0; j < (cnt); ++j) {
      int v;
      cin >> v;
      v--;
      E.emplace_back(i, v + M, A[i] + B[v]);
      ans += A[i] + B[v];
    }
  }
  sort((E).begin(), (E).end());
  reverse((E).begin(), (E).end());
  UnionFind uf(N + M);
  for (auto &e : E) {
    if (uf.merge(e.src, e.to)) {
      ans -= e.cost;
    }
  }
  cout << ans << ln;
}
