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.*;
import java.util.*;
import java.math.*;

public class E {
	static byte[] buf = new byte[1<<26];
    static int bp = -1;	
	
	public static void main(String[] args) throws IOException {

		/**/
		DataInputStream in = new DataInputStream(System.in);
		/*/
		DataInputStream in = new DataInputStream(new FileInputStream("src/e.in"));
		/**/
		
		in.read(buf, 0, 1<<26);
		int m = nni();
		int n = nni();
		int[] a = new int[m];
		int[] b = new int[n];
		for (int i = 0; i < m; ++i) {
			a[i] = nni();
		}
		for (int i = 0; i < n; ++i) {
			b[i] = nni();
		}
		int[][] sa = new int[m][];
		long tot = 0;
		TreeMap<Long, ArrayList<Long>> tm = new TreeMap<>();
		for (int i = 0; i < m; ++i) {
			sa[i] = new int[nni()];
			for (int j = 0; j < sa[i].length; ++j) {
				sa[i][j] = nni()-1;
				long cost = a[i]+b[sa[i][j]];
				tot += cost;
				if (!tm.containsKey(-cost))
					tm.put(-cost, new ArrayList<>());
				tm.get(-cost).add(((long)i)*1000000+sa[i][j]+m);
			}
		}
		int[] which = new int[n+m];
		ArrayList<ArrayList<Integer>> ch = new ArrayList<>();
		for (int i = 0; i < n+m; ++i) {
			ch.add(new ArrayList<>());
			which[i] = i;
			ch.get(i).add(i);
		}
		
		long ans = tot;
		for (long w : tm.keySet()) {
			for (long uv : tm.get(w)) {
				int u = (int)(uv/1000000);
				int v = (int)(uv%1000000);
				int pu = which[u];
				int pv = which[v];
				if (pu==pv)
					continue;
				ans += w;
				if (ch.get(pu).size() < ch.get(pv).size()) {
					int t = pv;
					pv = pu;
					pu = t;
				}
				for (int cv : ch.get(pv)) {
					which[cv] = pu;
					ch.get(pu).add(cv);
				}
				ch.get(pv).clear();
			}
		}
		System.out.println(ans);
	}
	
	public static int nni() {
        int ret = 0;
        byte b = buf[++bp];
        while (true) {
            ret = ret*10+b-'0';
            b = buf[++bp];
            if (b<'0'||b>'9') {
            	while (buf[bp+1]=='\r'||buf[bp+1]=='\n'||buf[bp+1]==' ') {++bp;}
            	break;
            }
        }
        return ret;
    }
}