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.
//package Graph.Spanning_Tree.Problems;

import java.util.Arrays;
import java.util.PriorityQueue;
import java.util.Scanner;

public class Avoid_Rainbow_Cycles {
    static long[] a, b;
    static long sum = 0;
    static int[] dis;
    static PriorityQueue<edge> edges;
    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        int m = scan.nextInt(), n = scan.nextInt();
        a = new long[m+1]; b = new long[n+1];
        for (int i = 1; i <= m; i++) a[i] = scan.nextLong();
        for (int i = 1; i <= n; i++) b[i] = scan.nextLong();

        // Maximum spanning tree
        edges = new PriorityQueue<>((e1, e2) -> (int) (e2.w - e1.w));
        dis = new int[m+n+5];
        Arrays.fill(dis, -1);
        for (int i = 1; i <= m; i++) {
            int sz = scan.nextInt();
            for (int j = 1; j <= sz; j++) {
                int num = scan.nextInt();
                edges.add(new edge(i, m + num, a[i] + b[num]));
                sum += a[i] + b[num];
            }
        }

        while (!edges.isEmpty()) {
            edge e = edges.poll();
            if (union(e.u, e.v)) sum -= e.w;
        }
        System.out.println(sum);
    }
    static int find(int x) {
        if (dis[x] < 0) return x;
        else return dis[x] = find(dis[x]);
    }
    static boolean union(int a, int b) {
        a = find(a); b = find(b);
        if (a == b) return false;
        if (dis[a] < dis[b]) {
            dis[a] += dis[b]; dis[b] = a;
        }else {
            dis[b] += dis[a]; dis[a] = b;
        }
        return true;
    }
    static class edge {
        int u, v;
        long w;
        edge(int u, int v, long w) {
            this.u = u; this.v = v;
            this.w = w;
        }
    }
}
