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;
const long long MOD = 1e9 + 7;
template <class T>
void chmin(T &a, const T &b) {
  if (a > b) a = b;
}
template <class T>
void chmax(T &a, const T &b) {
  if (a < b) a = b;
}
class DisjointSet {
 public:
  int N;
  vector<int> rank, p;
  vector<int> sz;
  DisjointSet() {}
  DisjointSet(int n) : N(n), rank(n), p(n), sz(n) {
    for (int i = 0; i < N; i++) makeSet(i);
  }
  void makeSet(int x) {
    p[x] = x;
    rank[x] = 0;
    sz[x] = 1;
  }
  bool same(int x, int y) { return leader(x) == leader(y); }
  void unite(int x, int y) {
    if (same(x, y)) return;
    link(leader(x), leader(y));
  }
  void link(int x, int y) {
    if (rank[x] > rank[y]) {
      p[y] = x;
      sz[x] += sz[y];
    } else {
      p[x] = y;
      sz[y] += sz[x];
      if (rank[x] == rank[y]) {
        ++rank[y];
      }
    }
  }
  int leader(int x) {
    if (x != p[x]) p[x] = leader(p[x]);
    return p[x];
  }
  int size(int x) { return sz[leader(x)]; }
};
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int M, N;
  cin >> M >> N;
  vector<long long> A(M);
  for (int i = 0; i < (M); ++i) cin >> A[i];
  vector<long long> B(N);
  for (int i = 0; i < (N); ++i) cin >> B[i];
  vector<vector<int>> C(M);
  for (int i = 0; i < (M); ++i) {
    int s;
    cin >> s;
    C[i].resize(s);
    for (int j = 0; j < (s); ++j) {
      cin >> C[i][j];
      C[i][j]--;
    }
  }
  vector<tuple<long long, int, int>> edges;
  long long all = 0;
  for (int i = 0; i < M; i++) {
    for (auto n : C[i]) {
      all += A[i] + B[n];
      edges.push_back(make_tuple(A[i] + B[n], i, M + n));
    }
  }
  sort(edges.begin(), edges.end());
  reverse(edges.begin(), edges.end());
  long long ans = 0;
  DisjointSet us(M + N);
  for (int i = 0; i < (edges.size()); ++i) {
    int a = get<1>(edges[i]);
    int b = get<2>(edges[i]);
    long long c = get<0>(edges[i]);
    if (us.same(a, b)) continue;
    us.unite(a, b);
    ans += c;
  }
  cout << all - ans << endl;
  return 0;
}
