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 a1, b1;
int a2, b2;
int k;
vector<int> d;
double f(int a1, int b1, int a2, int b2) {
  double ret = 0.0;
  for (int i = 0; i + k - 1 < d.size(); ++i) {
    if (d[i] < a1) continue;
    int tmp1 = max(a1, (i == 0 ? 0 : (d[i - 1] + 1)));
    int tmp2 = min(d[i], b1);
    int p = d[i + k - 1];
    int t1 = max(p, a2);
    int t2 = min(b2, max((i + k == d.size() ? b2 : (d[i + k] - 1)), a2 - 1));
    if (t1 <= t2)
      ret += (tmp2 - tmp1 + 1) / double(b1 - a1 + 1) * (t2 - t1 + 1) /
             double(b2 - a2 + 1);
    if (d[i] >= b1) break;
  }
  return ret;
}
int main() {
  scanf("%d%d%d%d%d", &a1, &b1, &a2, &b2, &k);
  for (int dig = 1; dig <= 9; ++dig)
    for (int m = 0; m < 1 << dig; ++m) {
      int num = 0;
      for (int i = 0; i < dig; ++i) {
        num *= 10;
        if ((m >> i) & 1)
          num += 7;
        else
          num += 4;
      }
      d.push_back(num);
    }
  sort(d.begin(), d.end());
  double res = 0.0;
  res += f(a1, b1, a2, b2);
  res += f(a2, b2, a1, b1);
  if (k == 1) {
    for (int i = 0; i < d.size(); ++i)
      if (d[i] >= a1 && d[i] <= b1 && d[i] >= a2 && d[i] <= b2)
        res -= 1 / double(b1 - a1 + 1) / double(b2 - a2 + 1);
  }
  printf("%.10lf\n", res);
  return 0;
}