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;
const double pi = acos(-1.0);
int a1, a2, b1, b2, k;
long long inter(int a, int b, int c, int d) {
  return max(0, min(b, d) - max(a, c) + 1);
}
int main() {
  scanf("%d%d%d%d%d", &a1, &a2, &b1, &b2, &k);
  vector<int> d;
  for (int _b = (9), len = (1); len <= _b; len++)
    for (int mask = 0, _n = (1 << len); mask < _n; mask++) {
      int x = 0;
      for (int i = 0, _n = (len); i < _n; i++)
        x = x * 10 + (mask & (1 << i) ? 4 : 7);
      d.push_back(x);
    }
  d.push_back(0);
  d.push_back(1000111000);
  sort((d).begin(), (d).end());
  long long res = 0;
  for (int i = 0, _n = ((int)(d).size()); i < _n; i++)
    if (i > 0) {
      if (i + k - 1 >= (int)(d).size() - 1) continue;
      res += inter(d[i - 1] + 1, d[i], a1, a2) *
             inter(d[i + k - 1], d[i + k] - 1, b1, b2);
      res += inter(d[i - 1] + 1, d[i], b1, b2) *
             inter(d[i + k - 1], d[i + k] - 1, a1, a2);
      if (k == 1 && a1 <= d[i] && d[i] <= a2 && b1 <= d[i] && d[i] <= b2) --res;
    }
  printf("%.12lf\n", res / (double)(a2 - a1 + 1) / (double)(b2 - b1 + 1));
  return 0;
}