Problem: Mahmoud wants to send a message to his friend Ehab. Their language consists of n words numbered from 1 to n. Some words have the same meaning so there are k groups of words such that all the words in some group have the same meaning.

Mahmoud knows that the i-th word can be sent with cost ai. For each word in his message, Mahmoud can either replace it with another word of the same meaning or leave it as it is. Can you help Mahmoud determine the minimum cost of sending the message?

The cost of sending the message is the sum of the costs of sending every word in it.

Input

The first line of input contains integers n, k and m (1 ≤ k ≤ n ≤ 105, 1 ≤ m ≤ 105) — the number of words in their language, the number of groups of words, and the number of words in Mahmoud's message respectively.

The second line contains n strings consisting of lowercase English letters of length not exceeding 20 which represent the words. It's guaranteed that the words are distinct.

The third line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109) where ai is the cost of sending the i-th word.

The next k lines describe the groups of words of same meaning. The next k lines each start with an integer x (1 ≤ x ≤ n) which means that there are x words in this group, followed by x integers which represent the indices of words in this group. It's guaranteed that each word appears in exactly one group.

The next line contains m space-separated words which represent Mahmoud's message. Each of these words appears in the list of language's words.

Output

The only line should contain the minimum cost to send the message after replacing some words (maybe none) with some words of the same meaning.

Examples

Input

5 4 4
i loser am the second
100 1 1 5 10
1 1
1 3
2 2 5
1 4
i am the second


Output

107

Input

5 4 4
i loser am the second
100 20 1 5 10
1 1
1 3
2 2 5
1 4
i am the second


Output

116

Note

In the first sample, Mahmoud should replace the word "second" with the word "loser" because it has less cost so the cost will be 100+1+5+1=107.

In the second sample, Mahmoud shouldn't do any replacement so the cost will be 100+1+5+10=116.

Solution (in PYTHON3):

Solution:
#Winners never quit, quiters never win............................................................................
from collections import deque as de
import math
from collections import Counter as cnt
from functools import reduce
from typing import MutableMapping
from itertools import groupby as gb

def factors(n):    
    return set(reduce(list.__add__, 
                ([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0)))
class My_stack():
    def __init__(self):
        self.data = []
    def my_push(self, x):
        return (self.data.append(x))
    def my_pop(self):
        return (self.data.pop())
    def my_peak(self):
        return (self.data[-1])
    def my_contains(self, x):
        return (self.data.count(x))
    def my_show_all(self):
        return (self.data)
    def isEmpty(self):
      return len(self.data)==0

arrStack = My_stack()    
def decimalToBinary(n): 
    return bin(n).replace("0b", "")

def isPrime(n) : 
	if (n <= 1) : 
		return False
	if (n <= 3) : 
		return True

	
	if (n % 2 == 0 or n % 3 == 0) : 
		return False

	i = 5
	while(i * i <= n) : 
		if (n % i == 0 or n % (i + 2) == 0) : 
			return False
		i = i + 6

	return True

def get_prime_factors(number):
    prime_factors = []
    while number % 2 == 0:
        prime_factors.append(2)
        number = number / 2
    for i in range(3, int(math.sqrt(number)) + 1, 2):
        while number % i == 0:
            prime_factors.append(int(i))
            number = number / i

    if number > 2:
        prime_factors.append(int(number))

    return prime_factors
def get_frequency(list):
    dic={}
    for ele in list:
        if ele in dic:
            dic[ele] += 1
        else:
            dic[ele] = 1
    return dic
def Log2(x): 
    return (math.log10(x) / 
            math.log10(2)); 

def isPowerOfTwo(n): 
    return (math.ceil(Log2(n)) == math.floor(Log2(n))); 

   
#here we go......................
#Winners never quit, Quitters never win
n,k,m=map(int,input().split())
wordindex = {}
for i, v in enumerate(input().split()):
    wordindex[int(i)+1] = v
    
costindex= {}
for i, v in enumerate(input().split()):
    costindex[int(i)+1] = int(v)
#print(wordindex)
#print(costindex)
grpcostdict={}
wordgrp={}
kk=k
while k:
    k-=1
    temp=list(map(int, input().split()))
    grpcostlist=[]
    for i in range(1,temp[0]+1):
        #print(temp[i],"cost")
        grpcostlist.append(costindex[temp[i]])
        wordgrp[wordindex[temp[i]]]=kk-k
        
    
    
    #print(kk-k,"kk-k")
    #print(grpcostlist,"grpcostlist")
    grpcostlist.sort()
    grpcostdict[kk-k]=grpcostlist
#print(grpcostdict)
#print(wordgrp)

message=list(map(str, input().split()))
#print(message)

ans=0
for i in range(m):
    ans+=grpcostdict[wordgrp[message[i]]][0]
print(ans)