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;
vector<int> vi;
void f(long long int x) {
  if (x > 1e9) return;
  vi.push_back(x);
  f(x * 10 + 4);
  f(x * 10 + 7);
}
int main() {
  f(4), f(7);
  double ans = 0;
  int vl, vr, pr, pl, k;
  cin >> pl >> pr >> vl >> vr >> k;
  sort(vi.begin(), vi.end());
  for (int i = 0; i < (int)vi.size(); i++) {
    int a, b = 1e9, x = 1, y;
    if (i + k - 1 < (int)vi.size()) {
      a = vi[i + k - 1];
      if (i + k < (int)vi.size()) b = vi[i + k] - 1;
      y = vi[i];
      if (i - 1 > -1) x = vi[i - 1] + 1;
      if (max(a, vl) <= min(b, vr) && max(x, pl) <= min(y, pr))
        ans += (min(b, vr) - max(a, vl) + 1.0) / (vr - vl + 1) *
               (min(y, pr) - max(x, pl) + 1.0) / (pr - pl + 1);
      if (max(a, pl) <= min(b, pr) && max(x, vl) <= min(y, vr))
        ans += (min(b, pr) - max(a, pl) + 1.0) / (pr - pl + 1) *
               (min(y, vr) - max(x, vl) + 1.0) / (vr - vl + 1);
    }
  }
  if (k == 1 && max(pl, vl) <= min(vr, pr)) {
    ans -= (1.0 / (pr - pl + 1) * 1.0 / (vr - vl + 1)) *
           (upper_bound(vi.begin(), vi.end(), min(vr, pr)) -
            lower_bound(vi.begin(), vi.end(), max(pl, vl)) + 0.0);
  }
  printf("%.10f\n", ans);
  return 0;
}