110_D. Lucky Probability
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.
```python
#!/usr/bin/env python3

vl, vr, pl, pr, k = map(int, input().split())
lucky = [4, 7]
sz = 0
for i in range(1, 9):
    base = 10 ** i
    psz, sz = sz, len(lucky)
    for j in [4 * base, 7 * base]:
        for pos in range(psz, sz):
            lucky.append(j + lucky[pos])

ans = 0
for i in range(0, len(lucky)-k+1):
    ll, lr = 1 if i == 0 else lucky[i-1] + 1, lucky[i]
    rl, rr = lucky[i+k-1], 1_000_000_000 if i+k == len(lucky) else lucky[i+k] - 1
    vc = max(0, min(vr, lr) - max(vl, ll) + 1)
    pc = max(0, min(pr, rr) - max(pl, rl) + 1)
    cnt1 = vc * pc
    vc = max(0, min(pr, lr) - max(pl, ll) + 1)
    pc = max(0, min(vr, rr) - max(vl, rl) + 1)
    cnt2 = vc * pc
    ans += cnt1 + cnt2
    if k == 1 and cnt1 > 0 and cnt2 > 0:
        ans -= 1

print(ans / ((vr - vl + 1) * (pr - pl + 1)))

```