Petya loves lucky numbers. Everybody knows that lucky numbers are positive integers whose decimal representation contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.

Let next(x) be the minimum lucky number which is larger than or equals x. Petya is interested what is the value of the expression next(l) + next(l + 1) + ... + next(r - 1) + next(r). Help him solve this problem.

Input

The single line contains two integers l and r (1 ≤ l ≤ r ≤ 109) — the left and right interval limits.

Output

In the single line print the only number — the sum next(l) + next(l + 1) + ... + next(r - 1) + next(r).

Please do not use the %lld specificator to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specificator.

Examples

Input

2 7


Output

33


Input

7 7


Output

7

Note

In the first sample: next(2) + next(3) + next(4) + next(5) + next(6) + next(7) = 4 + 4 + 4 + 7 + 7 + 7 = 33

In the second sample: next(7) = 7
import java.io.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.FileNotFoundException;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.LinkedList;
import java.util.Queue;
import java.util.StringTokenizer;


public class CF121A {
	static ArrayList<Long> op;
	static long max;
public static void main(String []args) throws FileNotFoundException {
      MyScanner in = new MyScanner();
      out = new PrintWriter(new BufferedOutputStream(System.out));
      //Start your solution from here.
      long a = in.nextLong();
      long b = in.nextLong();
      long ans = 0;
      max = b+1;
      op = new ArrayList<Long>();
      getlucky();
      for(int i=0;i<op.size();++i){
    	  if(op.get(i)>=a){
    		  long count = Math.min(b, op.get(i));
    		  ans += (long)(count - a + 1) * (op.get(i));
    		  a = count + 1;
    	  }
      }
      out.println(ans);
      //End your solution here.       
      out.close();
   }
 static void getlucky(){
	 Queue<String> q = new LinkedList<String>();
	 q.add("4");
	 q.add("7");
	 while(!q.isEmpty()){
		 String s = q.poll();
		 op.add(Long.parseLong(s));
		 if(Long.parseLong(s)>max){
			 break;
		 }
		 q.add(new String(s).concat("4"));
		 q.add(new String(s).concat("7"));
	 }
 }

   //-----------PrintWriter for faster output---------------------------------
   public static PrintWriter out;

   //-----------MyScanner class for faster input----------
   public static class MyScanner {
      BufferedReader br;
      StringTokenizer st;

      public MyScanner() {
         br = new BufferedReader(new InputStreamReader(System.in));
      }

      String next() {
          while (st == null || !st.hasMoreElements()) {
              try {
                  st = new StringTokenizer(br.readLine());
            } catch (IOException e) {
                  e.printStackTrace();
              }
          }
          return st.nextToken();
      }

      int nextInt() {
          return Integer.parseInt(next());
      }

      long nextLong() {
          return Long.parseLong(next());
      }

      double nextDouble() {
          return Double.parseDouble(next());
      }
      char nextChar()  {
    	  return next().charAt(0);
      }
      int [] nextIntAr(int [] ar){
    	  for(int i=0;i<ar.length;++i){
    		  ar[i] = Integer.parseInt(next());
    	  }
    	  return ar;
      }
      long [] nextLongAr(long [] ar){
    	  for(int i=0;i<ar.length;++i){
    		  ar[i] = Long.parseLong(next());
    	  }
    	  return ar;
      }
      double [] nextDoubleAr(double [] ar){
    	  for(int i=0;i<ar.length;++i){
    		  ar[i] = Double.parseDouble(next());
    	  }
    	  return ar;
      }
      String [] nextStringAr(String [] ar){
    	  for(int i=0;i<ar.length;++i){
    		  ar[i] = next();
    	  }
    	  return ar;
      }

      String nextLine(){
          String str = "";
      try {
         str = br.readLine();
      } catch (IOException e) {
         e.printStackTrace();
      }
      return str;
      }

   }
}