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;
long long k, l1, l2, r1, r2;
vector<long long> lucky_numbers, q[2];
void go(long long tam, long long num) {
  if (tam == 11) return;
  num *= 10;
  lucky_numbers.push_back(num + 4);
  lucky_numbers.push_back(num + 7);
  go(tam + 1, num + 4), go(tam + 1, num + 7);
}
long long Intersection(long long l1, long long r1, long long l2, long long r2) {
  return max(min(r1, r2) - max(l1, l2) + 1, 0LL);
}
int main() {
  cout << setprecision(15) << fixed;
  go(0, 0);
  lucky_numbers.push_back(0);
  sort(lucky_numbers.begin(), lucky_numbers.end());
  cin >> l1 >> r1 >> l2 >> r2 >> k;
  long long a, b, want = 0;
  for (int i = 0, j; i < (int)lucky_numbers.size() - k; ++i) {
    j = i + k;
    a = Intersection(lucky_numbers[i - 1] + 1, lucky_numbers[i], l1, r1);
    b = Intersection(lucky_numbers[j - 1], lucky_numbers[j] - 1, l2, r2);
    want += a * b;
    a = Intersection(lucky_numbers[i - 1] + 1, lucky_numbers[i], l2, r2);
    b = Intersection(lucky_numbers[j - 1], lucky_numbers[j] - 1, l1, r1);
    want += a * b;
    if (k == 1 && l1 <= lucky_numbers[i] && r1 >= lucky_numbers[i] &&
        l2 <= lucky_numbers[i] && r2 >= lucky_numbers[i])
      want--;
  }
  cout << want / (double)((r2 - l2 + 1) * (r1 - l1 + 1)) << endl;
  return 0;
}