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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int inf = 1 << 30;
set<long long> s;
void create(long long num, int sz) {
  if (sz >= 1 && sz <= 9) s.insert(num);
  if (sz <= 10) create(num * 10 + 4, sz + 1), create(num * 10 + 7, sz + 1);
}
long long x[20000];
int sz;
double izq(long long x, long long a[2]) {
  if (x < a[0]) return 0;
  return (min(a[1], x) - a[0] + 1.0);
}
double der(long long x, long long a[2]) {
  if (x > a[1]) return 0;
  return (a[1] - max(a[0], x) + 1.0);
}
int main() {
  create(0, 0);
  sz = 0;
  for (typeof s.begin() it = s.begin(); it != s.end(); it++) x[sz++] = *it;
  long long a[2], b[2];
  cin >> a[0] >> a[1];
  cin >> b[0] >> b[1];
  int k;
  cin >> k;
  double res = 0;
  for (int i = 0, j = k - 1; j < sz; j++, i++)
    res += (izq(x[i], a) - (i == 0 ? 0.0 : izq(x[i - 1], a))) *
           (der(x[j], b) - (j + 1 == sz ? 0.0 : der(x[j + 1], b))),
        res += (izq(x[i], b) - (i == 0 ? 0.0 : izq(x[i - 1], b))) *
               (der(x[j], a) - (j + 1 == sz ? 0.0 : der(x[j + 1], a)));
  if (k == 1)
    for (int i = 0; i < sz; i++)
      if (a[0] <= x[i] && x[i] <= a[1] && b[0] <= x[i] && x[i] <= b[1]) res--;
  printf("%.010f\n", res / (a[1] - a[0] + 1) / (b[1] - b[0] + 1));
}