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.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.List;
import java.util.StringTokenizer;
import java.io.BufferedReader;
import java.util.Comparator;
import java.util.Collections;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        EAvoidRainbowCycles solver = new EAvoidRainbowCycles();
        solver.solve(1, in, out);
        out.close();
    }

    static class EAvoidRainbowCycles {
        public void solve(int testNumber, InputReader in, PrintWriter out) {
            int m = in.nextInt(), n = in.nextInt();
            int[] a = in.readIntArray(m);
            int[] b = in.readIntArray(n);
            List<Graphs.Edge> edges = new ArrayList<>();
            for (int i = 0; i < m; i++) {
                int s = in.nextInt();
                for (int j = 0; j < s; j++) {
                    int u = in.nextInt() - 1;
                    edges.add(new Graphs.Edge(i, m + u, a[i] + b[u]));
                }
            }
            Collections.sort(edges, Comparator.comparingInt(e -> -e.cost));
            UnionFind uf = new UnionFind(n + m);
            long answer = 0;
            for (Graphs.Edge e : edges) {
                if (uf.rep(e.i) == uf.rep(e.j)) {
                    answer += e.cost;
                    continue;
                }
                uf.union(e.i, e.j);
            }

            out.println(answer);
        }

    }

    static class InputReader {
        public final BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

        public int[] readIntArray(int n) {
            int[] x = new int[n];
            for (int i = 0; i < n; i++) {
                x[i] = nextInt();
            }
            return x;
        }

    }

    static class Graphs {
        public static class Edge {
            public final int i;
            public final int j;
            public final int cost;

            public Edge(int i, int j, int cost) {
                this.i = i;
                this.j = j;
                this.cost = cost;
            }

            public String toString() {
                return String.format("[(%d, %d): %d]", i, j, cost);
            }

        }

    }

    static class UnionFind {
        private final int[] __rep;
        private final int[] __size;

        public UnionFind(int n) {
            __rep = new int[n];
            __size = new int[n];
            for (int i = 0; i < n; i++) {
                __rep[i] = i;
                __size[i] = 1;
            }
        }

        public int rep(int x) {
            if (__rep[x] == x) {
                return x;
            }

            int r = rep(__rep[x]);
            __rep[x] = r;
            return r;
        }

        public int size(int x) {
            return __size[rep(x)];
        }

        public boolean union(int x, int y) {
            x = rep(x);
            y = rep(y);

            if (x == y) {
                return false;
            }

            if (size(x) < size(y)) {
                int t = x;
                x = y;
                y = t;
            }

            // now size(x) >= size(y)

            __rep[y] = x;
            __size[x] += __size[y];
            return true;
        }

    }
}

