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;
using ll = int64_t;
using ld = long double;
using P = pair<int, int>;
using vs = vector<string>;
using vi = vector<int>;
using vvi = vector<vi>;
template <class T>
using PQ = priority_queue<T>;
template <class T>
using PQG = priority_queue<T, vector<T>, greater<T> >;
const int INF = 0xccccccc;
const ll LINF = 0xcccccccccccccccLL;
template <typename T1, typename T2>
inline bool chmax(T1 &a, T2 b) {
  return a < b && (a = b, true);
}
template <typename T1, typename T2>
inline bool chmin(T1 &a, T2 b) {
  return a > b && (a = b, true);
}
template <typename T1, typename T2>
istream &operator>>(istream &is, pair<T1, T2> &p) {
  return is >> p.first >> p.second;
}
template <typename T1, typename T2>
ostream &operator<<(ostream &os, const pair<T1, T2> &p) {
  return os << p.first << ' ' << p.second;
}
struct UnionFind {
  vector<int> par;
  UnionFind(int n) : par(n, -1) {}
  int root(int x) {
    if (par[x] < 0) return x;
    return par[x] = root(par[x]);
  }
  bool unite(int x, int y) {
    x = root(x);
    y = root(y);
    if (x == y) return false;
    if (par[x] > par[y]) swap(x, y);
    par[x] += par[y];
    par[y] = x;
    return true;
  }
  bool same(int x, int y) { return root(x) == root(y); }
  int size(int x) { return -par[root(x)]; }
};
struct edge {
  int u, v;
  int64_t cost;
  edge(int u = 0, int v = 0, int64_t cost = 0) : u(u), v(v), cost(cost) {}
};
bool comp(const edge &e1, const edge &e2) { return e1.cost > e2.cost; }
vector<edge> es;
int64_t kruskal(int V) {
  int E = es.size();
  sort(es.begin(), es.end(), comp);
  UnionFind uft(V);
  int64_t res = 0;
  for (int i = 0; i < E; ++i) {
    edge e = es[i];
    if (!uft.same(e.u, e.v)) {
      uft.unite(e.u, e.v);
      res += e.cost;
    }
  }
  return res;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int m, n;
  cin >> m >> n;
  vi a(m), b(n);
  for (int i = 0; i < m; ++i) cin >> a[i];
  for (int i = 0; i < n; ++i) cin >> b[i];
  ll ans = 0;
  for (int i = 0; i < m; ++i) {
    int s;
    cin >> s;
    for (int j = 0; j < s; ++j) {
      int z;
      cin >> z;
      es.emplace_back(z - 1, i + n, a[i] + b[z - 1]);
      ans += a[i] + b[z - 1];
    }
  }
  cout << ans - kruskal(n + m) << endl;
}
