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 int inf = 1e9 + 7;
const long long longinf = 1LL << 60;
const long long mod = 1e9 + 7;
const long double eps = 1e-10;
template <typename T1, typename T2>
inline void chmin(T1& a, T2 b) {
  if (a > b) a = b;
}
template <typename T1, typename T2>
inline void chmax(T1& a, T2 b) {
  if (a < b) a = b;
}
class UnionFind {
 public:
  int par[200050];
  int depth[200050];
  int nGroup[200050];
  UnionFind(int n) { init(n); }
  void init(int n) {
    for (int i = 0; i < n; i++) {
      par[i] = i;
      depth[i] = 0;
      nGroup[i] = 1;
    }
  }
  int root(int x) {
    if (par[x] == x) {
      return x;
    } else {
      return par[x] = root(par[x]);
    }
  }
  bool same(int x, int y) { return root(x) == root(y); }
  void unite(int x, int y) {
    x = root(x);
    y = root(y);
    if (x == y) return;
    if (depth[x] < depth[y]) {
      par[x] = y;
      nGroup[y] += nGroup[x];
      nGroup[x] = 0;
    } else {
      par[y] = x;
      nGroup[x] += nGroup[y];
      nGroup[y] = 0;
      if (depth[x] == depth[y]) depth[x]++;
    }
  }
};
struct edge {
  int u;
  int v;
  long long cost;
  edge(int _u, int _v, long long _cost) : u(_u), v(_v), cost(_cost) {}
};
bool comp(const edge& e1, const edge& e2) { return e1.cost < e2.cost; }
vector<pair<int, int>> conn;
long long kruskal(int N, vector<edge>& es) {
  sort(es.begin(), es.end(), comp);
  UnionFind uf(N);
  long long res = 0;
  for (int i = (int)(0); i < (int)(es.size()); i++) {
    edge e = es[i];
    if (!uf.same(e.u, e.v)) {
      uf.unite(e.u, e.v);
      res += e.cost;
      conn.push_back(make_pair(e.u, e.v));
    }
  }
  return res;
}
void solve() {
  int m, n;
  cin >> m >> n;
  vector<long long> A(m);
  for (int i = (int)(0); i < (int)(m); i++) cin >> A[i];
  vector<long long> B(n);
  for (int i = (int)(0); i < (int)(n); i++) cin >> B[i];
  vector<edge> edges;
  long long sum = 0;
  for (int i = (int)(0); i < (int)(m); i++) {
    int sz;
    cin >> sz;
    for (int j = (int)(0); j < (int)(sz); j++) {
      int k;
      cin >> k;
      k--;
      edges.push_back({i, m + k, -(A[i] + B[k])});
      sum += A[i] + B[k];
    }
  }
  auto res = kruskal(n + m, edges);
  cout << sum + res << '\n';
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int T = 1;
  while (T--) solve();
  return 0;
}
