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;
vector<string> vec_splitter(string s) {
  s += ',';
  vector<string> res;
  while (!s.empty()) {
    res.push_back(s.substr(0, s.find(',')));
    s = s.substr(s.find(',') + 1);
  }
  return res;
}
void debug_out(vector<string> __attribute__((unused)) args,
               __attribute__((unused)) int idx,
               __attribute__((unused)) int LINE_NUM) {
  cerr << endl;
}
template <typename Head, typename... Tail>
void debug_out(vector<string> args, int idx, int LINE_NUM, Head H, Tail... T) {
  if (idx > 0)
    cerr << ", ";
  else
    cerr << "Line(" << LINE_NUM << ") ";
  stringstream ss;
  ss << H;
  cerr << args[idx] << " = " << ss.str();
  debug_out(args, idx + 1, LINE_NUM, T...);
}
const int LOG = 17;
const int NN = 1 << LOG;
const int MAXN = (int)60000 + 5;
const int INF = 1 << 29;
const long long MAX = (long long)1e18 + 5;
const int MOD = 1000;
long long mul(long long x, long long y) { return (1LL * x * y) % MOD; }
long long add(long long x, long long y) {
  long long res = x;
  res += y;
  return (res % MOD + MOD) % MOD;
}
int two(int x) { return 1 << x; }
bool contain(int mask, int bit) { return (mask & two(bit)) > 0; }
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
int bit_count(int x) { return x == 0 ? 0 : (x & 1) + bit_count(x >> 1); }
int n, m;
int a[NN], b[NN];
struct Edge {
  int x, y, w;
  Edge(int _x, int _y, int _w) {
    this->x = _x;
    this->y = _y;
    this->w = _w;
  }
  friend bool operator<(const Edge& A, const Edge& B) { return A.w > B.w; }
};
int h[NN * 2];
int par[NN * 2];
int root(int x) { return par[x] = (x == par[x] ? x : root(par[x])); }
bool join(int x, int y) {
  x = root(x), y = root(y);
  if (x == y) return false;
  if (h[x] > h[y]) {
    par[y] = x;
  } else {
    par[x] = y;
    if (h[x] == h[y]) h[y]++;
  }
  return true;
}
int main() {
  ios_base::sync_with_stdio(false);
  scanf("%d%d", &m, &n);
  for (int i = 0; i < m; i++) scanf("%d", &a[i]);
  for (int i = 0; i < n; i++) scanf("%d", &b[i]);
  vector<Edge> edges;
  long long ans = 0;
  for (int i = 0; i < m; i++) {
    int s;
    scanf("%d", &s);
    while (s--) {
      int j;
      scanf("%d", &j);
      j--;
      edges.push_back(Edge(i, j + m, a[i] + b[j]));
      ans += a[i] + b[j];
    }
  }
  sort(edges.begin(), edges.end());
  memset(h, 0, sizeof(h));
  for (int i = 0; i < n + m; i++) par[i] = i;
  for (Edge& e : edges) {
    if (join(e.x, e.y)) {
      ans -= e.w;
    }
  }
  printf("%lld\n", ans);
}
