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:
import sys

def genForLen(n):
    res = []
    for i in xrange(1 << n):
        s = 0
        for j in xrange(n):
            s = s * 10 + (4 if ((i >> j) & 1) != 0 else 7)
        res.append(s)
    return res


def genLucky(n):
    res = []
    for i in xrange(n):
        res += genForLen(i + 1)
    return res

def calcStrong(pl, pr, vl, vr, preFirst, first, last, postLast, equalIsGood):
    preFirst = max(preFirst, pl)
    first = min(first, pr)
    last = max(last, vl)
    postLast = min(postLast, vr)
    a = first - preFirst + 1
    b = postLast - last + 1
    if a <= 0 or b <= 0:
        return 0
    res = a * b
    if not equalIsGood and first == last:
        res -= 1

    return res

def countGood(pl, pr, vl, vr, k):
    nums = genLucky(9)
    nums.sort()
    res = 0L

    for i in xrange(len(nums) - k + 1):
        first, last = nums[i], nums[i + k - 1]
        preFirst = (0 if i == 0 else nums[i - 1]) + 1
        postLast = (1000000001 if i + k == len(nums) else nums[i + k]) - 1
        res += calcStrong(pl, pr, vl, vr, preFirst, first, last, postLast, True)
        res += calcStrong(vl, vr, pl, pr, preFirst, first, last, postLast, False)

    return res

pl, pr, vl, vr, k = map(int, sys.stdin.readline().split())

res = countGood(pl, pr, vl, vr, k) / (float(pr - pl + 1) * (vr - vl + 1))
print "%.10f" % res