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 A>
ostream &operator<<(ostream &cout, vector<A> const &v);
template <typename A, typename B>
ostream &operator<<(ostream &cout, pair<A, B> const &p) {
  return cout << "(" << p.first << ", " << p.second << ")";
}
template <typename A>
ostream &operator<<(ostream &cout, vector<A> const &v) {
  cout << "[";
  for (int i = 0; i < v.size(); i++) {
    if (i) cout << ", ";
    cout << v[i];
  }
  return cout << "]";
}
template <typename A, typename B>
istream &operator>>(istream &cin, pair<A, B> &p) {
  cin >> p.first;
  return cin >> p.second;
}
mt19937_64 rng(std::chrono::steady_clock::now().time_since_epoch().count());
void usaco(string filename) {
  freopen((filename + ".in").c_str(), "r", stdin);
  freopen((filename + ".out").c_str(), "w", stdout);
}
const long double pi = 3.14159265358979323846;
const long long mod = 1000000007;
struct dsu {
  int n;
  vector<int> sizes, marks;
  vector<vector<int>> graph;
  void init(int rn) {
    n = rn;
    sizes = vector<int>(n);
    fill(sizes.begin(), sizes.end(), 1);
    marks = vector<int>(n);
    for (int i = 0; i < n; i++) marks[i] = i;
    graph = vector<vector<int>>(n);
  }
  void mark(int m, int node) {
    if (marks[node] == m) return;
    marks[node] = m;
    for (int i : graph[node]) mark(m, i);
  }
  bool merge(int a, int b) {
    if (marks[a] == marks[b]) return 0;
    graph[a].push_back(b);
    graph[b].push_back(a);
    if (sizes[marks[a]] > sizes[marks[b]]) swap(a, b);
    sizes[marks[b]] += sizes[marks[a]];
    sizes[marks[a]] = 0;
    mark(marks[b], a);
    return 1;
  }
  int find(int n) { return marks[n]; }
};
long long n, m, k, q, l, r, x, y, z;
const long long template_array_size = 1e6 + 6884;
long long a[template_array_size];
long long b[template_array_size];
long long c[template_array_size];
string second, t;
long long ans = 0;
pair<long long, pair<long long, long long>> vals[200005];
long long root[100005];
bool cnt[100005];
dsu d;
void solve(int tc = 0) {
  cin >> m >> n;
  for (long long i = 0; i < m; i++) cin >> a[i];
  for (long long i = 0; i < n; i++) cin >> b[i];
  long long ans = 0, second = 0;
  for (long long i = 0; i < m; i++) {
    cin >> x;
    for (long long j = 0; j < x; j++) {
      cin >> y;
      --y;
      ans += a[i] + b[y];
      vals[second++] = make_pair(-a[i] - b[y], make_pair(i, y));
    }
  }
  sort(vals, vals + second);
  memset(cnt, 0, sizeof(cnt));
  d.init(n);
  for (long long i = 0; i < second; i++) {
    long long a = vals[i].second.first, b = vals[i].second.second,
              val = vals[i].first;
    if (!cnt[a]) {
      cnt[a] = 1;
      root[a] = b;
      ans += val;
    } else {
      if (d.merge(root[a], b)) ans += val;
    }
  }
  cout << ans << '\n';
}
int main() {
  { ios_base::sync_with_stdio(false); }
  { cin.tie(NULL); }
  cout << setprecision(12) << fixed;
  int tc = 1;
  for (int t = 0; t < tc; t++) solve(t);
}
