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.util.List;
import java.util.stream.LongStream;
import java.util.stream.Stream;
import java.util.Collection;
import java.io.IOException;
import java.util.ArrayList;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) throws Exception {
        Thread thread = new Thread(null, new TaskAdapter(), "", 1 << 29);
        thread.start();
        thread.join();
    }

    static class TaskAdapter implements Runnable {
        @Override
        public void run() {
            InputStream inputStream = System.in;
            OutputStream outputStream = System.out;
            FastInput in = new FastInput(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, FastInput in, PrintWriter out) {
            int m = in.readInt();
            int n = in.readInt();

            int[] a = new int[m];
            int[] b = new int[n];
            in.populate(a);
            in.populate(b);

            int g = n + m;

            List<Edge> edges = new ArrayList<>();
            for (int i = 0; i < m; i++) {
                int s = in.readInt();
                for (int j = 0; j < s; j++) {
                    int x = in.readInt() - 1;
                    edges.add(new Edge(x, n + i, a[i] + b[x]));
                }
            }

            edges.sort((x, y) -> -Long.compare(x.cost, y.cost));
            DSU dsu = new DSU(g);
            dsu.reset();

            long ans = edges.stream().mapToLong(x -> x.cost).sum();
            for (Edge e : edges) {
                if (dsu.find(e.a) != dsu.find(e.b)) {
                    dsu.merge(e.a, e.b);
                    ans -= e.cost;
                }
            }

            out.println(ans);
        }

    }

    static class FastInput {
        private final InputStream is;
        private byte[] buf = new byte[1 << 13];
        private int bufLen;
        private int bufOffset;
        private int next;

        public FastInput(InputStream is) {
            this.is = is;
        }

        public void populate(int[] data) {
            for (int i = 0; i < data.length; i++) {
                data[i] = readInt();
            }
        }

        private int read() {
            while (bufLen == bufOffset) {
                bufOffset = 0;
                try {
                    bufLen = is.read(buf);
                } catch (IOException e) {
                    bufLen = -1;
                }
                if (bufLen == -1) {
                    return -1;
                }
            }
            return buf[bufOffset++];
        }

        public void skipBlank() {
            while (next >= 0 && next <= 32) {
                next = read();
            }
        }

        public int readInt() {
            int sign = 1;

            skipBlank();
            if (next == '+' || next == '-') {
                sign = next == '+' ? 1 : -1;
                next = read();
            }

            int val = 0;
            if (sign == 1) {
                while (next >= '0' && next <= '9') {
                    val = val * 10 + next - '0';
                    next = read();
                }
            } else {
                while (next >= '0' && next <= '9') {
                    val = val * 10 - next + '0';
                    next = read();
                }
            }

            return val;
        }

    }

    static class Edge {
        int a;
        int b;
        long cost;

        public Edge(int a, int b, long cost) {
            this.a = a;
            this.b = b;
            this.cost = cost;
        }

    }

    static class DSU {
        protected int[] p;
        protected int[] rank;

        public DSU(int n) {
            p = new int[n];
            rank = new int[n];
        }

        public void reset() {
            for (int i = 0; i < p.length; i++) {
                p[i] = i;
                rank[i] = 0;
            }
        }

        public final int find(int a) {
            if (p[a] == p[p[a]]) {
                return p[a];
            }
            preAccess(a);
            return p[a] = find(p[a]);
        }

        protected void preAccess(int a) {

        }

        protected void preMerge(int a, int b) {

        }

        public final void merge(int a, int b) {
            a = find(a);
            b = find(b);
            if (a == b) {
                return;
            }
            if (rank[a] == rank[b]) {
                rank[a]++;
            }

            if (rank[a] < rank[b]) {
                int tmp = a;
                a = b;
                b = tmp;
            }

            preMerge(a, b);
            p[b] = a;
        }

    }
}

