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 time,math as mt,bisect as bs,sys
from sys import stdin,stdout
from collections import deque
from fractions import Fraction
from collections import Counter
from collections import OrderedDict
pi=3.14159265358979323846264338327950
def II(): # to take integer input
    return int(stdin.readline())
def IP(): # to take tuple as input
    return map(int,stdin.readline().split())
def L(): # to take list as input
    return list(map(int,stdin.readline().split()))
def P(x): # to print integer,list,string etc..
    return stdout.write(str(x)+"\n")
def PI(x,y): # to print tuple separatedly
    return stdout.write(str(x)+" "+str(y)+"\n")
def lcm(a,b): # to calculate lcm
    return (a*b)//gcd(a,b)
def gcd(a,b): # to calculate gcd
    if a==0:
        return b
    elif b==0:
        return a
    if a>b:
        return gcd(a%b,b)
    else:
        return gcd(a,b%a)
def bfs(adj,v): # a schema of bfs
    visited=[False]*(v+1)
    q=deque()
    while q:
        pass
def sieve():
    li=[True]*(2*(10**5)+5)
    li[0],li[1]=False,False
    for i in range(2,len(li),1):
        if li[i]==True:
            for j in range(i*i,len(li),i):
                li[j]=False
    prime=[]
    for i in range((2*(10**5)+5)):
        if li[i]==True:
            prime.append(i)
    return prime
def setBit(n):
    count=0
    while n!=0:
        n=n&(n-1)
        count+=1
    return count
mx=10**7
spf=[mx]*(mx+1)
def SPF():
    spf[1]=1
    for i in range(2,mx+1):
        if spf[i]==mx:
            spf[i]=i
            for j in range(i*i,mx+1,i):
                if i<spf[j]:
                    spf[j]=i
    return
def readTree(n,e): # to read tree
    adj=[set() for i in range(n+1)]
    for i in range(e):
        u1,u2=IP()
        adj[u1].add(u2)
    return adj
#####################################################################################
mod=10**9+7
def get(s,li):
    if s!="":
        li.append(int(s))
        if int(s)>1e10:
            return
    get(s+"4",li)
    get(s+"7",li)
    return

def solve():
    l,r=IP()
    li=[]
    get("",li)
    li.insert(l,0)
    li.sort()
    ans = 0
    for x in li:
        if x >= r:
            ans += x*(r-l+1)
            break
        elif x >= l:
            ans += x*(x-l+1)
            l = x+1
    print(ans)
    return
t=1
for i in range(t):
    solve()

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