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 static java.lang.Integer.parseInt;
import static java.lang.Long.parseLong;
import static java.lang.System.arraycopy;
import static java.lang.System.exit;
import static java.util.Arrays.copyOf;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.NoSuchElementException;
import java.util.StringTokenizer;

public class E {

	static class IntList {

		int data[] = new int[3];
		int size = 0;

		boolean isEmpty() {
			return size == 0;
		}

		int size() {
			return size;
		}

		int get(int index) {
			if (index < 0 || index >= size) {
				throw new IndexOutOfBoundsException();
			}
			return data[index];
		}

		void clear() {
			size = 0;
		}

		void set(int index, int value) {
			if (index < 0 || index >= size) {
				throw new IndexOutOfBoundsException();
			}
			data[index] = value;
		}

		void expand() {
			if (size >= data.length) {
				data = copyOf(data, (data.length << 1) + 1);
			}
		}

		void insert(int index, int value) {
			if (index < 0 || index > size) {
				throw new IndexOutOfBoundsException();
			}
			expand();
			arraycopy(data, index, data, index + 1, size++ - index);
			data[index] = value;
		}

		int delete(int index) {
			if (index < 0 || index >= size) {
				throw new IndexOutOfBoundsException();
			}
			int value = data[index];
			arraycopy(data, index + 1, data, index, --size - index);
			return value;
		}

		void push(int value) {
			expand();
			data[size++] = value;
		}

		int pop() {
			if (size == 0) {
				throw new NoSuchElementException();
			}
			return data[--size];
		}

		void unshift(int value) {
			expand();
			arraycopy(data, 0, data, 1, size++);
			data[0] = value;
		}

		int shift() {
			if (size == 0) {
				throw new NoSuchElementException();
			}
			int value = data[0];
			arraycopy(data, 1, data, 0, --size);
			return value;
		}
	}

	static void sortBy(int a[], int n, int v[]) {
		if (n == 0) {
			return;
		}
		for (int i = 1; i < n; i++) {
			int j = i;
			int ca = a[i];
			int cv = v[ca];
			do {
				int nj = (j - 1) >> 1;
				int na = a[nj];
				if (cv <= v[na]) {
					break;
				}
				a[j] = na;
				j = nj;
			} while (j != 0);
			a[j] = ca;
		}
		int ca = a[0];
		for (int i = n - 1; i > 0; i--) {
			int j = 0;
			while ((j << 1) + 2 + Integer.MIN_VALUE < i + Integer.MIN_VALUE) {
				j <<= 1;
				j += (v[a[j + 2]] > v[a[j + 1]]) ? 2 : 1;
			}
			if ((j << 1) + 2 == i) {
				j = (j << 1) + 1;
			}
			int na = a[i];
			a[i] = ca;
			ca = na;
			int cv = v[ca];
			while (j != 0 && v[a[j]] < cv) {
				j = (j - 1) >> 1;
			}
			while (j != 0) {
				na = a[j];
				a[j] = ca;
				ca = na;
				j = (j - 1) >> 1;
			}
		}
		a[0] = ca;
	}

	static int dsuGet(int dsu[], int i) {
		return dsu[i] == i ? i : (dsu[i] = dsuGet(dsu, dsu[i]));
	}

	static boolean dsuMerge(int dsu[], int i, int j) {
		i = dsuGet(dsu, i);
		j = dsuGet(dsu, j);
		if (i == j) {
			return false;
		}
		dsu[j] = i;
		return true;
	}

	static void solve() throws Exception {
		int m = scanInt(), n = scanInt();
		int a[] = new int[m], b[] = new int[n];
		for (int i = 0; i < m; i++) {
			a[i] = scanInt();
		}
		for (int i = 0; i < n; i++) {
			b[i] = scanInt();
		}
		IntList ia = new IntList(), ib = new IntList();
		for (int i = 0; i < m; i++) {
			int s = scanInt();
			for (int j = 0; j < s; j++) {
				ia.push(i);
				ib.push(scanInt() - 1);
			}
		}
		int nn = ia.size;
		int cost[] = new int[nn], idx[] = new int[nn];
		for (int i = 0; i < nn; i++) {
			cost[i] = a[ia.data[i]] + b[ib.data[i]];
			idx[i] = i;
		}
		sortBy(idx, nn, cost);
		int dsu[] = new int[m + n];
		for (int i = 0; i < m + n; i++) {
			dsu[i] = i;
		}
		long ans = 0;
		for (int i = nn - 1; i >= 0; i--) {
			if (!dsuMerge(dsu, ia.data[idx[i]], m + ib.data[idx[i]])) {
				ans += cost[idx[i]];
			}
		}
		out.print(ans);
	}

	static int scanInt() throws IOException {
		return parseInt(scanString());
	}

	static long scanLong() throws IOException {
		return parseLong(scanString());
	}

	static String scanString() throws IOException {
		while (tok == null || !tok.hasMoreTokens()) {
			tok = new StringTokenizer(in.readLine());
		}
		return tok.nextToken();
	}

	static BufferedReader in;
	static PrintWriter out;
	static StringTokenizer tok;

	public static void main(String[] args) {
		try {
			in = new BufferedReader(new InputStreamReader(System.in));
			out = new PrintWriter(System.out);
			solve();
			in.close();
			out.close();
		} catch (Throwable e) {
			e.printStackTrace();
			exit(1);
		}
	}
}