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.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.StringTokenizer;

public class C {
    private static BufferedReader in;
    private static StringTokenizer st;
    private static PrintWriter out;
    
    public static void main(String[] args) throws NumberFormatException, IOException {
        in = new BufferedReader(new InputStreamReader(System.in));
        st = new StringTokenizer("");
        out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
        int count = 0;
        ArrayList<Integer> list = new ArrayList<Integer>();
        long lucky[]=  new long[2000];
        int cnt = 0;
        int a[] = new int [10];
        for (int i = 0; i < (1<<9); i++) {
            int t = i;
            cnt = 0;
            Arrays.fill(a, 0);
            while(t > 0){
                a[cnt++] = t % 2;
                t/=2;
            }
            int next1=0;
            int next2=0;
            for (int j = 0; j < cnt; j++) {
                if(a[j]==0){
                    next1 = next1 * 10 + 4;
                }else{
                    next1 = next1 * 10 + 7;
                }
                if(a[j]==0){
                    next2 = next2 * 10 + 7;
                }else{
                    next2 = next2 * 10 + 4;
                }
            }
            if(cnt > 0){
                lucky[count++] = next1;
                lucky[count++] = next2;
            }
        }
        lucky[count++] = Long.parseLong("4444444444"); 
        Arrays.sort(lucky,0,count);
        long l = nextInt();
        long r = nextInt();
        int ind1 = Arrays.binarySearch(lucky, 0,count,l);
        if(ind1 < 0){
            ind1 = -ind1 - 1;
        }
        int ind2 = Arrays.binarySearch(lucky, 0,count,r);
        if(ind2 < 0){
            ind2 = -ind2 - 1;
        }
        long ans = 0;
        for (int i = ind1; i < ind2; i++) {
            ans += (long) (lucky[i]-l + 1) * lucky[i];
            l = lucky[i] + 1;
        }
        ans += (long) (r - l + 1) * lucky[ind2];
        System.out.println(ans);
    }

    static String next() throws IOException{
        while(!st.hasMoreTokens()){
            st = new StringTokenizer(in.readLine());
        }
        return st.nextToken();
    }
    static int nextInt() throws NumberFormatException, IOException{
        return Integer.parseInt(next());
    }
    static long nextLong() throws NumberFormatException, IOException{
        return Long.parseLong(next());
    }
    
    static double nextDouble() throws NumberFormatException, IOException{
        return Double.parseDouble(next());
    }

}
