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 GraknE {
    static ArrayList<Edge> edges;
    public static void main(String[] args) {
        MyScanner sc = new MyScanner();
        PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
        int m = sc.nextInt(); int n = sc.nextInt();
        long [] a = new long[m];
        long [] b = new long[n];
        edges = new ArrayList<>();
        for (int i = 0; i < m; i++) a[i] = sc.nextInt();
        for (int i = 0; i < n; i++) b[i] = sc.nextInt();
        for (int i = 0; i < m; i++) {
            int sz = sc.nextInt();
            for (int j = 0; j < sz; j++) {
                int node = sc.nextInt() - 1;
                edges.add(new Edge(node, i + n, a[i] + b[node]));
            }
        }
        UF dsu = new UF(n + m);
        Collections.sort(edges, Comparator.comparingLong(x -> x.w));
        long total = 0;
        for (int i = 0; i < edges.size(); i++) total -= edges.get(i).w;
        long res = 0; int mst = 0; int cur = 0;
        while (mst < m + n - 1 && cur < edges.size()) {
            if (!dsu.connected(edges.get(cur).u, edges.get(cur).v)) {
                dsu.union(edges.get(cur).u, edges.get(cur).v);
                ++mst;
                res += edges.get(cur).w;
            }
            ++cur;
        }
        out.println(total + res);
        out.close();
    }



    static class Edge {
        int u; int v; long w;
        Edge(int u, int v, long w) {
            this.w = -w; this. u = u; this.v = v;
        }
    }

    static class UF {
        private int[] parent;  // parent[i] = parent of i
        private byte[] rank;   // rank[i] = rank of subtree rooted at i (never more than 31)
        private int count;     // number of components
        private int[] size;
        public UF(int n) {
            if (n < 0) throw new IllegalArgumentException();
            count = n;
            parent = new int[n];
            rank = new byte[n];
            size = new int[n];
            for (int i = 0; i < n; i++) {
                parent[i] = i;
                rank[i] = 0;
                size[i] = 1;
            }
        }

        public int find(int p) {
            while (p != parent[p]) {
                parent[p] = parent[parent[p]];    // path compression by halving
                p = parent[p];
            }
            return p;
        }

        public int count() {
            return count;
        }

        public boolean connected(int p, int q) {
            return find(p) == find(q);
        }

        public void union(int p, int q) {
            int rootP = find(p);
            int rootQ = find(q);
            if (rootP == rootQ) return;

            // make root of smaller rank point to root of larger rank
            if (rank[rootP] < rank[rootQ]) {
                parent[rootP] = rootQ;
                size[rootQ] = size[rootQ] + size[rootP];
            }
            else if (rank[rootP] > rank[rootQ]) {
                parent[rootQ] = rootP;
                size[rootP] = size[rootP] + size[rootQ];
            }
            else {
                parent[rootQ] = rootP;
                size[rootP] = size[rootP] + size[rootQ];
                rank[rootP]++;
            }
            count--;
        }
    }


    //-----------MyScanner class for faster input----------
    public static class MyScanner {
        BufferedReader br;
        StringTokenizer st;

        public MyScanner() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

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

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }


    }

}