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;
void __print(int x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T &x) {
  int f = 0;
  cerr << '{';
  for (auto &i : x) cerr << (f++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
inline long long int GCD(long long int x, long long int y) {
  if (x < y) swap(x, y);
  if (x == 0) return y;
  if (y == 0) return x;
  return GCD(x % y, y);
}
long long int phi(long long int n) {
  long long int result = n;
  for (long long int i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      while (n % i == 0) n /= i;
      result -= result / i;
    }
  }
  if (n > 1) result -= result / n;
  return result;
}
long long int power(long long int x, long long int n, long long int mod) {
  long long int res = 1;
  x %= mod;
  while (n) {
    if (n & 1) {
      res = ((res * x) % mod + mod) % mod;
    }
    x = ((x * x) % mod + mod) % mod;
    n >>= 1;
  }
  return res;
}
const int MAXN = 1e5 + 5;
long long int N, M;
long long int A[MAXN], B[MAXN];
class DSU {
 public:
  long long int N;
  vector<long long int> parent, sz;
  DSU(long long int N) {
    this->N = N + 5;
    parent.assign(N + 2, 0);
    sz.assign(N + 2, 1);
    for (long long int i = 1; i <= N; i++) parent[i] = i;
  }
  long long int find_parent(long long int x) {
    while (x != parent[x]) {
      x = parent[parent[x]];
    }
    return x;
  }
  bool merge(long long int x, long long int y) {
    long long int a = find_parent(x);
    long long int b = find_parent(y);
    if (a == b) return false;
    if (sz[a] < sz[b]) swap(a, b);
    sz[a] += sz[b];
    parent[b] = parent[a];
    return true;
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> M >> N;
  for (long long int i = 0; i < M; i++) {
    cin >> A[i];
  }
  for (long long int i = 0; i < N; i++) {
    cin >> B[i];
  }
  vector<pair<long long int, pair<long long int, long long int> > > edges;
  long long int k, x;
  for (long long int i = 0; i < M; i++) {
    cin >> k;
    for (long long int j = 0; j < k; j++) {
      cin >> x;
      edges.push_back(make_pair(A[i] + B[x - 1], make_pair(x, i + N + 1)));
    }
  }
  sort(edges.begin(), edges.end());
  reverse(edges.begin(), edges.end());
  DSU dsu(N + M + 5);
  long long int ans = 0;
  for (long long int i = 0; i < edges.size(); i++) {
    if (!dsu.merge(edges[i].second.first, edges[i].second.second)) {
      ans += edges[i].first;
    }
  }
  cout << ans << endl;
}
