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.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author lucasr
 */
public class Main {
	public static void main(String[] args) {
		InputStream inputStream = System.in;
		OutputStream outputStream = System.out;
		MyScanner in = new MyScanner(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, MyScanner sc, PrintWriter out) {
			int m = sc.nextInt();
			int n = sc.nextInt();
			long[] setCost = sc.nextLongArray(m);
			long[] elemCost = sc.nextLongArray(n);
			long total = 0;
			List<EAvoidRainbowCycles.Edge> edges = new ArrayList<>();
			for (int i = 0; i < m; i++) {
				int size = sc.nextInt();
				for (int j = 0; j < size; j++) {
					int elem = sc.nextInt() - 1;
					long cost = setCost[i] + elemCost[elem];
					total += cost;
					edges.add(new EAvoidRainbowCycles.Edge(i, elem, cost));
				}
			}
			UnionFind uf = new UnionFind(m + n);
			edges.sort((e1, e2) -> Long.compare(e2.cost, e1.cost));
			for (EAvoidRainbowCycles.Edge edge : edges) {
				int a = uf.find(edge.set);
				int b = uf.find(edge.elem + m);
				if (a != b) {
					uf.union(a, b);
					total -= edge.cost;
				}
			}
			out.println(total);
		}

		static class Edge {
			int set;
			int elem;
			long cost;

			public Edge(int set, int elem, long cost) {
				this.set = set;
				this.elem = elem;
				this.cost = cost;
			}

		}

	}

	static class UnionFind {
		int[] rep;

		public UnionFind(int size) {
			rep = new int[size];
			for (int i = 0; i < size; i++) {
				rep[i] = i;
			}
		}

		public int find(int val) {
			if (rep[val] != val) {
				return rep[val] = find(rep[val]);
			}
			return val;
		}

		public boolean union(int a, int b) {
			int aRep = find(a);
			int bRep = find(b);
			if (aRep == bRep) return false;
			rep[aRep] = bRep;
			return true;
		}

	}

	static class MyScanner {
		private BufferedReader br;
		private StringTokenizer tokenizer;

		public MyScanner(InputStream is) {
			br = new BufferedReader(new InputStreamReader(is));
		}

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

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

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

		public long[] nextLongArray(int n) {
			long[] ret = new long[n];
			for (int i = 0; i < n; i++) {
				ret[i] = nextLong();
			}
			return ret;
		}

	}
}

