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:
/* Author: Ronak Agarwal, reader part not written by me*/
import java.io.* ; import java.util.* ;
import static java.lang.Math.min ; import static java.lang.Math.max ;
import static java.lang.Math.abs ; import static java.lang.Math.log ;
import static java.lang.Math.pow ; import static java.lang.Math.sqrt ;
/* Thread is created here to increase the stack size of the java code so that recursive dfs can be performed */
public class Contest{
	public static void main(String[] args) throws IOException{
		new Thread(null,new Runnable(){	public void run(){ exec_Code() ;} },"Solver",1l<<27).start() ;
	}
	static void exec_Code(){
		try{
			Solver Machine = new Solver() ;
			Machine.Solve() ; Machine.Finish() ;}
		catch(Exception e){ e.printStackTrace() ; print_and_exit() ;}
		catch(Error e){ e.printStackTrace() ; print_and_exit() ; }
	}
	static void print_and_exit(){ System.out.flush() ; System.exit(-1) ;}
}
/* Implementation of the Reader class */
class Reader { 
	private InputStream mIs;
	private byte[] buf = new byte[1024];
	private int curChar,numChars;
	public Reader() { this(System.in); } 
	public Reader(InputStream is) { mIs = is;} 
	public int read() {
		if (numChars == -1) throw new InputMismatchException();    
		if (curChar >= numChars) {
			curChar = 0;
			try { numChars = mIs.read(buf);} catch (IOException e) { throw new InputMismatchException();}
			if (numChars <= 0) return -1;            
		}
		return buf[curChar++];
	} 
	public String nextLine(){
		int c = read();
		while (isSpaceChar(c)) c = read();        
		StringBuilder res = new StringBuilder();
		do {
			res.appendCodePoint(c);
			c = read();
		} while (!isEndOfLine(c));
		return res.toString() ;
	} 
	public String s(){
		int c = read();
		while (isSpaceChar(c)) c = read();        
		StringBuilder res = new StringBuilder();
		do {
			res.appendCodePoint(c);
			c = read();
		}while (!isSpaceChar(c));
		return res.toString();
	} 
	public long l(){
		int c = read();
		while (isSpaceChar(c)) c = read();        
		int sgn = 1;
		if (c == '-') { sgn = -1 ; c = read() ; }
		long res = 0;
		do{
			if (c < '0' || c > '9') throw new InputMismatchException();
			res *= 10 ; res += c - '0' ; c = read();
		}while(!isSpaceChar(c));
		return res * sgn;
	} 
	public int i(){
		int c = read() ;
		while (isSpaceChar(c)) c = read();        
		int sgn = 1;
		if (c == '-') { sgn = -1 ; c = read() ; }
		int res = 0;
		do{
			if (c < '0' || c > '9') throw new InputMismatchException();
			res *= 10 ; res += c - '0' ; c = read() ;
		}while(!isSpaceChar(c));
		return res * sgn;
	} 
	public boolean isSpaceChar(int c) { return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1; } 
	public boolean isEndOfLine(int c) { return c == '\n' || c == '\r' || c == -1; } 
}   
/* All the useful functions,constants,renamings are here*/
class Template{
	/* Constants Section */
	final int INT_MIN = Integer.MIN_VALUE ; final int INT_MAX = Integer.MAX_VALUE ;
	final long LONG_MIN = Long.MIN_VALUE ; final long LONG_MAX = Long.MAX_VALUE ;
	static long MOD = 1000000007 ;
	static Reader ip = new Reader() ;	
	static PrintWriter op = new PrintWriter(System.out) ;
	/* Methods for writing */
	static void p(Object o){ op.print(o) ; }
	static void pln(Object o){ op.println(o) ;}
	static void Finish(){ op.flush(); op.close(); }
	/* Implementation of operations modulo MOD */
	static long inv(long a){ return powM(a,MOD-2) ; }
	static long m(long a,long b){ return (a*b)%MOD ; }
	static long d(long a,long b){ return (a*inv(b))%MOD ; }
	static long powM(long a,long b){
		if(b<0) return powM(inv(a),-b) ;
		long ans=1 ;
		for( ; b!=0 ; a=(a*a)%MOD , b/=2) if((b&1)==1) ans = (ans*a)%MOD ;
		return ans ;
	}
	/* Renaming of some generic utility classes */
	final static class mylist extends ArrayList<Integer>{}
	final static class myset extends TreeSet<Integer>{}
	final static class mystack extends Stack<Integer>{}
	final static class mymap extends TreeMap<Integer,Integer>{}
}
/* Implementation of the pair class, useful for every other problem */
class pair implements Comparable<pair>{
	int x ; int y ;
	pair(int _x,int _y){ x=_x ; y=_y ;} 
	public int compareTo(pair p){
		return (this.x<p.x ? -1 : (this.x>p.x ? 1 : (this.y<p.y ? -1 : (this.y>p.y ? 1 : 0)))) ;
	}
}
/* Main code starts here */
class Solver extends Template{
	public void Solve() throws IOException{
		int n = ip.i() ; int k = ip.i() ; int m = ip.i() ;
		String data[] = new String[n+1] ;
		for(int i=1 ; i<=n ; i++)
			data[i] = ip.s() ;
		long a[] = new long[n+1] ;
		for(int i=1 ; i<=n ; i++) a[i] = ip.i() ;
		int grp[] = new int[n+1] ;
		for(int i=1 ; i<=k ; i++){
			int x = ip.i() ;
			for(int j=1 ; j<=x ; j++) grp[ip.i()] = i ;
		}
		long mng[] = new long[k+1] ;
		Arrays.fill(mng,LONG_MAX) ;
		for(int i=1 ; i<=n ; i++){
			int j = grp[i] ;
			mng[j] = min(mng[j],a[i]) ;
		}
		HashMap<String,Integer> mp = new HashMap<String,Integer>() ;
		for(int i=1 ; i<=n ; i++) mp.put(data[i],i) ;
		long res=0 ;
		for(int i=1 ; i<=m ; i++)
			res+=mng[grp[mp.get(ip.s())]] ;
		pln(res) ;
 	}
}