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> lucky;
void func0(int x) {
  if (x > 0) lucky.push_back(x);
  for (int i = 4; i < 8; i += 3)
    if (10LL * x + i < 1000000000) func0(10 * x + i);
}
int func(int l, int r, int a, int b) {
  if (l <= a && b <= r) return b - a + 1;
  if (l <= a) return max(0, r - a + 1);
  if (b <= r) return max(0, b - l + 1);
  return r - l + 1;
}
int main() {
  func0(0);
  lucky.push_back(0);
  lucky.push_back(2000000000);
  sort(lucky.begin(), lucky.end());
  int pl, pr, vl, vr, k;
  while (~scanf("%d%d%d%d%d", &pl, &pr, &vl, &vr, &k)) {
    double a = (pr - pl + 1.) * (vr - vl + 1.);
    double res = 0;
    for (int i = (1); i < (lucky.size() - k); ++i) {
      int p0 = func(pl, pr, lucky[i - 1] + 1, lucky[i]);
      int v0 = func(vl, vr, lucky[i - 1] + 1, lucky[i]);
      int p1 = func(pl, pr, lucky[i + k - 1], lucky[i + k] - 1);
      int v1 = func(vl, vr, lucky[i + k - 1], lucky[i + k] - 1);
      double a = (1. * p0 * v1) / (pr - pl + 1.);
      double b = (1. * p1 * v0) / (pr - pl + 1.);
      res += a + b;
      if (k == 1 && a > 0 && b > 0) res -= 1. / (pr - pl + 1.);
    }
    printf("%.10f\n", res / (vr - vl + 1.));
  }
  return 0;
}