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 JAVA):

Solution:
//package practice_codes;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.Map.Entry;
import java.util.PriorityQueue;
import java.util.stream.Collectors;

//import codes.Reference.i;

public class MahmoudAndEhabAndTheMessage {
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		
		long t[];
		
		t = i.la();
		long k = t[1];
		long m = t[2];
		String s[] = i.sa();
		long cost[] = i.la();
		HashMap<String, Integer> map = new HashMap<String, Integer>();
		for(int i=0;i<s.length;i++)
		{
			map.put(s[i], i);
		}
		HashMap<Long, Long> map1 = new HashMap<Long, Long>();
		while(k>0)
		{
			long a[]=i.la();
			if(a[0]>1)
			{
				long min = Long.MAX_VALUE;
				long ind=0;
				for(int i=1;i<a.length;i++)
				{
					map1.put(a[i]-1, a[i]-1);
					if(cost[(int)(a[i]-1)]<min)
					{
						min = cost[(int)(a[i]-1)];
						ind=a[i]-1;
					}
				}
				for(int i=1;i<a.length;i++)
				{
					map1.put(a[i]-1, ind);
					
				}
				
			}
			k--;	
		}
		String sa[]=i.sa();
		long sum=0;
		for(int i=0;i<sa.length;i++)
		{
			if(map1.containsKey((long)map.get(sa[i])))
			{
				long ndx = map1.get((long)map.get(sa[i]));
				sum+=cost[(int)ndx];
			}
			else
			{
				sum+=cost[map.get(sa[i])];
			}
		}
		System.out.println(sum);
	}
	static private class i
	{
		//input taking methods
		static BufferedReader b = new BufferedReader(new InputStreamReader(System.in));
		
		public static String[] sa()
	    {
	    	try {
				return b.readLine().split(" ");
			} catch (IOException e) {
				
				e.printStackTrace();
			}
	    	return null;
	    }
		
		public static long l()
		{
			try {
				return Long.parseLong(b.readLine());
			} catch (IOException e) {
				// TODO Auto-generated catch block
				e.printStackTrace();
			}
	    	return 0;
		}
		
		public static String s()
		{
			try {
				return b.readLine();
			} catch (IOException e) {
				// TODO Auto-generated catch block
				e.printStackTrace();
			}
	    	return null;
		}
		
		public static long[] la()
		{
			return Arrays.stream(sa()).parallel().mapToLong(Long::parseLong).toArray();
		}
		
		public static ArrayList<Long> lal()
		{
			return (ArrayList<Long>)Arrays.stream(la()).parallel().boxed().collect(Collectors.toList());
		}
		public static StringBuilder so(String a)
		{
			return new StringBuilder(a);
		}

	}
	/*
	 HashMap<String,Integer> map = new HashMap<String,Integer>();
	 if(!map.containsKey(key))
	 {
	 	map.put(key,value);
	 }
	 else
	 {
	 	map.replace(key,map.get(key));
	 } 
	 */
	static private class algo{
		
		public static boolean isPrime(int a)
		{
			for(int i=2;i<a;i++)
			{
				if(a%i==0)
				{
					return false;
				}
			}
			return true;
		}
		static String reverse(String s)
		{
			return i.so(s).reverse().toString();
		}
		static String replace(String s,int st,int e,String repl)
		{
			return i.so(s).replace(st, e, repl).toString();
		}
		static String insert(String s,int pos,String ins)
		{
			return i.so(s).insert(pos, ins).toString();
		}
		static long sod(long n) 
	    { 
	        long sum; 
	        for (sum = 0; n > 0; sum += n % 10, 
	                                  n /= 10); 
	  
	        return sum; 
	    } 

		static int gcd(int a, int b) 
	    { 
			if (a == 0) 
				return b;  
			return gcd(b % a, a);  
	    } 
		
	    static int lcm(int a, int b) 
	    { 
	        return (a*b)/gcd(a, b); 
	    }
	    static long sumOfSquares(long[] a)
	    {
	    	
	    	return (long) Arrays.stream(a).mapToDouble(i->i*i).sum();
	    }
	    //use of filter
	    //Arrays.stream(a).filter(k->k==1).count();
	    //            array       variable->condition returning boolean  . lot fof functionsa re available
	    
	    static String[] generateBinSeq(int cnt)
	    {
	    	int loopvar=(int) Math.pow(2, cnt);
			String a[][] = new String[cnt+1][loopvar];
			
			for(int i=1;i<=cnt;i++)
			{
				int temp = (int) Math.pow(2, i);
				for(int j=0;j<temp;j++)
				{
					if(i==1 || i==2)
					{
						if(i==1)
						{
							a[i][0]="0";
							a[i][1]="1";
							continue;
						}
						else
						{
							a[2][0]="00";a[2][1]="01";a[2][2]="10";a[2][3]="11";
							continue;
						}
					}
					else
					{
						if(j<(temp/2))
						{
							a[i][j] = a[i-1][j]+"0"; 
						}
						else
						{
							a[i][j] = a[i-1][j%((temp+1)/2)]+"1";
						}
				
					}
						
				}
			}
	    
			return a[cnt];
	    }
	    
	 // USe this snippet for calling
//	    int ndx = algo.bs(a,key,0,a.length-1);
//		if(ndx<0)
//			ndx=0;
//		if(ndx==a.length)
//			ndx=a.length-1;

	    static public int bs(long a[],long key,int s,int e)
		{
			int m = (s+e)/2;
			if(s>e)
				return s;
				
			if(a[m]==key)
				return m;
			else if(a[m]>key)
				return bs(a,key,s,m-1);
			else
				return bs(a,key,m+1,e);
			
		}
	 // USe this snippet for calling
//	    int ndx = bsa(al,key,0,al.size()-1);
//		if(ndx<0)
//			ndx=0;
//		if(ndx==al.size())
//			ndx=al.size()-1;
	    static public int bsa(ArrayList<Long> a,long key,int s,int e)
		{
			int m = (s+e)/2;
			if(s>e)
				return s;
				
			if(a.get(m)==key)
				return m;
			else if(a.get(m)>key)
				return bsa(a,key,s,m-1);
			else
				return bsa(a,key,m+1,e);
			
		}
	    static public long fact(long a)
	    {
	    	long i,fact=1;  
	    	 long number=a;   
	    	  for(i=1;i<=number;i++){    
	    	      fact=fact*i;    
	    	  } 
	    	  return fact;
	    }
	    
	}
}