Problem: Petya loves lucky numbers. We all know that lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.

Petya and his friend Vasya play an interesting game. Petya randomly chooses an integer p from the interval [pl, pr] and Vasya chooses an integer v from the interval [vl, vr] (also randomly). Both players choose their integers equiprobably. Find the probability that the interval [min(v, p), max(v, p)] contains exactly k lucky numbers.

Input

The single line contains five integers pl, pr, vl, vr and k (1 ≤ pl ≤ pr ≤ 109, 1 ≤ vl ≤ vr ≤ 109, 1 ≤ k ≤ 1000).

Output

On the single line print the result with an absolute error of no more than 10 - 9.

Examples

Input

1 10 1 10 2


Output

0.320000000000


Input

5 6 8 10 1


Output

1.000000000000

Note

Consider that [a, b] denotes an interval of integers; this interval includes the boundaries. That is, <image>

In first case there are 32 suitable pairs: (1, 7), (1, 8), (1, 9), (1, 10), (2, 7), (2, 8), (2, 9), (2, 10), (3, 7), (3, 8), (3, 9), (3, 10), (4, 7), (4, 8), (4, 9), (4, 10), (7, 1), (7, 2), (7, 3), (7, 4), (8, 1), (8, 2), (8, 3), (8, 4), (9, 1), (9, 2), (9, 3), (9, 4), (10, 1), (10, 2), (10, 3), (10, 4). Total number of possible pairs is 10·10 = 100, so answer is 32 / 100.

In second case Petya always get number less than Vasya and the only lucky 7 is between this numbers, so there will be always 1 lucky number.

Solution (in PYTHON):

Solution:
pl,pr,vl,vr,k=map(int,raw_input().split())
def conv(x):
  t=0
  while x>1:
    t=t*10+(7 if x&1 else 4)
    x>>=1
  return t
a=filter(lambda x: x>=min(pl,vl) and x<=max(pr,vr),[conv(x) for x in xrange(2,1<<11)])
a.sort()
n,s=len(a),0
a+=[10**10]
def size(a,b,c,d):
  return max(0,min(b,d)-max(a,c)+1)
for i in xrange(n-k+1):
  p1=a[i-1]+1 if i>0 else 1
  p2=a[i]
  v1,v2=a[i+k-1],a[i+k]-1
  s+=size(p1,p2,pl,pr)*size(v1,v2,vl,vr)
  s+=size(p1,p2,vl,vr)*size(v1,v2,pl,pr)
  if 1==k and pl<=p2<=pr and vl<=p2<=vr:s-=1
print s/1.0/(pr-pl+1)/(vr-vl+1)