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.
import java.util.*;
import java.io.*;

public class CFA {
  BufferedReader br;
  PrintWriter out;
  StringTokenizer st;
  boolean eof;
  private static final long MOD = 1000L * 1000L * 1000L + 7;
  private static final int[] dx = {0, -1, 0, 1};
  private static final int[] dy = {1, 0, -1, 0};
  private static final String yes = "Yes";
  private static final String no = "No";

  void solve() {
//    int T = nextInt();
    int T = 1;
    for (int i = 0; i < T; i++) {
      helper();
    }
  }

  void helper() {
    int m = nextInt();
    int n = nextInt();
    int[] a = nextIntArr(m);
    int[] b = nextIntArr(n);
    List<Edge> ls = new ArrayList<>();
    for (int i = 0; i < m; i++) {
      int cnt = nextInt();
      for (int j = 0; j < cnt; j++) {
        int val = nextInt() - 1;
        ls.add(new Edge(i, m + val, a[i] + b[val]));
      }
    }
    Collections.sort(ls, new Comparator<Edge>() {
      @Override
      public int compare(Edge o1, Edge o2) {
        return Integer.compare(o2.weight, o1.weight);
      }
    });

    long res = 0;
    WeightedUF uf = new WeightedUF(m + n);
    for (Edge e : ls) {
      if (uf.isConnected(e.from, e.to)) {
        res += e.weight;
        continue;
      }
      uf.union(e.from, e.to);
    }
    outln(res);
  }

  class Edge {
    int from;
    int to;
    int weight;
    public Edge(int from, int to, int weight) {
      this.from = from;
      this.to = to;
      this.weight = weight;
    }
  }

  // Weighted Union Find.
  class WeightedUF {
    int[] id;
    int count;
    int[] sz;
    public WeightedUF(int n) {
      count = n;
      id = new int[n];
      for(int i = 0; i < n; i++) {
        id[i] = i;
      }
      sz = new int[n];
      Arrays.fill(sz, 1);
    }

    public boolean isConnected(int p, int q) {
      return find(p) == find(q);
    }
    public int find(int p) {
      while(p != id[p]) {
        p = id[p];
      }
      return p;
    }

    public boolean union(int p, int q) {
      int i = find(p);
      int j = find(q);
      if(i == j) {
        return false;
      }
      if(sz[i] < sz[j]) {
        id[i] = j;
        sz[j] += sz[i];
      }
      else {
        id[j] = i;
        sz[i] += sz[j];
      }
      count--;
      return true;
    }
  }

  void shuffle(int[] a) {
    int n = a.length;
    for(int i = 0; i < n; i++) {
      int r = i + (int) (Math.random() * (n - i));
      int tmp = a[i];
      a[i] = a[r];
      a[r] = tmp;
    }
  }

  long gcd(long a, long b) {
    while(a != 0 && b != 0) {
      long c = b;
      b = a % b;
      a = c;
    }
    return a + b;
  }
  int gcd(int a, int b) {
    while(a != 0 && b != 0) {
      int c = b;
      b = a % b;
      a = c;
    }
    return a + b;
  }
  private void outln(Object o) {
    out.println(o);
  }
  private void out(Object o) {
    out.print(o);
  }
  private void formatPrint(double val) {
    outln(String.format("%.9f", val));
  }
  public CFA() {
    br = new BufferedReader(new InputStreamReader(System.in));
    out = new PrintWriter(System.out);
    solve();
    out.close();
  }
  public static void main(String[] args) {
    new CFA();
  }

  public long[] nextLongArr(int n) {
    long[] res = new long[n];
    for(int i = 0; i < n; i++)
      res[i] = nextLong();
    return res;
  }
  public int[] nextIntArr(int n) {
    int[] res = new int[n];
    for(int i = 0; i < n; i++)
      res[i] = nextInt();
    return res;
  }
  public String nextToken() {
    while (st == null || !st.hasMoreTokens()) {
      try {
        st = new StringTokenizer(br.readLine());
      } catch (Exception e) {
        eof = true;
        return null;
      }
    }
    return st.nextToken();
  }
  public String nextString() {
    try {
      return br.readLine();
    } catch (IOException e) {
      eof = true;
      return null;
    }
  }
  public int nextInt() {
    return Integer.parseInt(nextToken());
  }
  public long nextLong() {
    return Long.parseLong(nextToken());
  }
  public double nextDouble() {
    return Double.parseDouble(nextToken());
  }
}