// O(length^3 * letters^2) DP solution to 'Assembly line'.
// Enrique Martin Martin

import java.lang.*;
import java.util.*;
import java.io.*;
import java.util.regex.Pattern;

public class cheap{

   static final int MAXL = 200;             //Maximum chain length
   static final int MAXA = 'z'-'a' + 1;     //Maximum number of letters: 'z'-'a'
   static final int INF  = 1000000000;  

   static char letter[] = new char[MAXA];           //Positions of letters
   static int multCost[][] = new int[MAXA][MAXA];      //Cost of the multiplication of 2 letters
   static int multRes[][] = new int[MAXA][MAXA];     //Result of the multiplication of 2 letters
   static int cost[][][] = new int[MAXL][MAXL][MAXA];  //Minimum cost of multiplying in a substring cost[i][j] obtaining a particular letter
   static String s;   
   
   static private int getPosition( char c )
   {
      for( int i = 0; i < MAXA; ++i )
         if( letter[i] == c )
            return i;
         
      return -1; //This will not happen because c is supposed to appear in letter[]
   }                        

   static private void minimumCost( String s, int len, int numLetters )
   {
      int c1, c2, minc, actc;
      
      for( int i = 0; i < len; i++ )
        for( int j = 0; j < len; j++ )
          for( int k = 0; k < numLetters; k++ )
            cost[i][j][k] = INF;
     
      /*Base case, substrings of length 1*/
      for( int i = 0; i < len; ++i )
      {
         int pos = getPosition( s.charAt(i) );
         cost[i][i][pos] = 0;
      }
      
      /*Substrings of length > 1*/
      for( int diag = 1; diag < len; diag++ )  
      {
         for( int i = 0; i+diag < len; i++ )
         {
            /*Substring [i,i+diag]*/
            for( int k = i; k < i+diag; k++ )
            {
               for( int l1 = 0; l1 < numLetters; l1++ )
               {
                 for( int l2 = 0; l2 < numLetters; l2++ )
                 {             
                   if( cost[i][i+diag][ multRes[l1][l2] ] > (cost[i][k][l1] + cost[k+1][i+diag][l2] + multCost[l1][l2] ) )
                     cost[i][i+diag][ multRes[l1][l2] ] = cost[i][k][l1] + cost[k+1][i+diag][l2] + multCost[l1][l2];              
                 }
               }
            }
         }
      }
   }


   public static void main(String[] args)
   {
      int k, costMul, n, m, len;
      char res;
      Scanner scanner = new Scanner(System.in);
      
      m = 0;
      for(;;)
      {
         k = scanner.nextInt();

         if( k == 0 ) 
           break;
           
         if( m > 0 )
            System.out.println();
         m++;
         
         
         Pattern patPar = Pattern.compile("([0-9]+)-[a-z]");
         Pattern patChar = Pattern.compile("[a-z]");         
         Pattern patSpace = Pattern.compile("[ ]");
         
         for( int i = 0; i < k; i++ )
         {             
            letter[i] = scanner.next( patChar ).charAt(0);
         }
         
         for( int i = 0; i < k; i++ )
         {
            for( int j = 0; j < k; j++ )
            {          
              String par = scanner.next( patPar );
              costMul = Integer.parseInt( par.substring(0, par.length()-2) );
              res = par.charAt(par.length()-1);
              
              multRes[i][j] = getPosition(res);
              multCost[i][j] = costMul;              
            }
         }
         
         n = scanner.nextInt();
         scanner.nextLine();
         for( int i = 0; i < n; i++ )
         {
            s = scanner.nextLine();
            len = s.length();
            
            minimumCost( s, len, k );
            
            int mincost = INF;
            for( int j = 0; j < k; j++ )
            {
               if ( cost[0][len-1][j] < mincost )
                  mincost = cost[0][len-1][j];
            }
            
            for( int j = 0; j < k; j++ )
            {
               if ( cost[0][len-1][j] == mincost )
               {
                 System.out.print(mincost);
                 System.out.print("-");
                 System.out.println(letter[j]);
                 break;
               }
            } 
         }   
      }
   }

}