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.IOException;
import java.io.InputStreamReader;
import java.util.*;

public class LuckySum {
    static BufferedReader in;
    static StringTokenizer stk;
    static String next() {
        try {
            while (stk == null || !stk.hasMoreTokens())
                stk = new StringTokenizer(in.readLine());

            return stk.nextToken();
        } catch(Exception ex) {
            throw new RuntimeException("Failed reading input", ex);
        }
    }
    static int nextInt() { return Integer.parseInt(next()); }

    public static void main(String [] args) throws IOException {
        in = new BufferedReader(new InputStreamReader(System.in));

        int L = nextInt(), R = nextInt();
        luckyNums = generateLuckyNumbers();
        //luckyNums.forEach(System.out::println);
        //System.out.println(1000000000);
        //System.out.println(luckyNums.size());
        //System.out.println(sumInterval(40, 44));
        //System.out.println(nextLucky(800));

        long nextLucky = nextLucky(L);
        long from = L, to = Math.min(nextLucky, R);
        long total = 0;
        while(to<=R) {
            total += sumInterval(from, to, nextLucky);
            if(to==R) break;
            nextLucky = nextLucky(to+1);
            from = to+1;
            to = nextLucky > R ? R : nextLucky;
        }




        System.out.println(total);
    }

    static List<Long> luckyNums;

    static List<Long> generateLuckyNumbers() {
        List<Long> list = new ArrayList<>(1000);
        for(int length=1; length<10; length++) {
            luckyNumbersOfLength(length, list);
        }
        list.add(4444444444L);
        return list;
    }

    static void luckyNumbersOfLength(int length, List<Long> out) {
        final int limit = (1<<length);
        for(int mask=0; mask<limit; mask++) {
            out.add(toLuckyNumber(mask, length));
        }
    }


    static long toLuckyNumber(int mask, int length) {
        long output = 0;
        for(int pos=length-1; pos>=0; pos--) {
            int flag = (mask>>pos) & 1;
            int digit = (flag==0) ? 4 : 7;
            output = (output*10) + digit;
        }

        return output;
    }

    static long sumInterval(long from, long to, long luckyLimit) {
        long range = to - from + 1;
        return range*luckyLimit;
    }

    static long nextLucky(long nonLucky) {
        int index = Collections.binarySearch(luckyNums, nonLucky);
        if(index<0) index = -index -1;
        //System.out.println(index);

        return luckyNums.get(index);
    }
}