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 pl, pr, vl, vr;
int k;
long long Inter(long long x1, long long x2, long long y1, long long y2) {
  long long a = max(x1, y1);
  long long b = min(x2, y2);
  if (a > b)
    return 0.0;
  else
    return b - a + 1;
}
long long GetProb(long long x1, long long x2, long long y1, long long y2) {
  return Inter(x1, x2, pl, pr) * Inter(y1, y2, vl, vr) +
         Inter(x1, x2, vl, vr) * Inter(y1, y2, pl, pr);
}
int main() {
  cin >> pl >> pr >> vl >> vr >> k;
  vector<long long> numbers;
  numbers.push_back(0);
  int a = 0;
  for (int i = 0; i < 9; i++) {
    int b = numbers.size();
    for (int j = a; j < b; j++) {
      numbers.push_back(numbers[j] * 10 + 4);
      numbers.push_back(numbers[j] * 10 + 7);
    }
    a = b;
  }
  numbers.push_back(1000000000);
  long long P = (pr - pl + 1) * (vr - vl + 1);
  long long gP = 0;
  int l = 1;
  int r = k;
  while (1) {
    gP +=
        GetProb(numbers[l - 1] + 1, numbers[l], numbers[r], numbers[r + 1] - 1);
    l++;
    r++;
    if (r + 1 >= numbers.size()) break;
  }
  l = 1;
  r = k + 1;
  while (1) {
    l++;
    r++;
    if (r + 1 >= numbers.size()) break;
  }
  l = 1;
  if (k == 1)
    while (1) {
      gP -= GetProb(numbers[l], numbers[l], numbers[l], numbers[l]) / 2;
      l++;
      if (l + 1 >= numbers.size()) break;
    }
  printf("%.15f", (long double)gP / P);
  return 0;
}