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 p1, p2, v1, v2, k;
int main() {
  cin >> p1 >> p2 >> v1 >> v2 >> k;
  vector<int> res[10];
  res[1].push_back(4);
  res[1].push_back(7);
  for (int i = 2; i < 10; i++) {
    for (int j = 0; j < res[i - 1].size(); j++) {
      res[i].push_back(res[i - 1][j] * 10 + 4);
      res[i].push_back(res[i - 1][j] * 10 + 7);
    }
  }
  for (int i = 1; i < 10; i++) {
    res[0].insert(res[0].begin(), res[i].begin(), res[i].end());
  }
  sort(res[0].begin(), res[0].end());
  double ret = 0;
  if (k == 1) {
    for (int i = 0; i < res[0].size(); i++) {
      int l2 = v1;
      int r2 = v2;
      int l1 = p1;
      if (i > 0) {
        l1 = max(res[0][i - 1] + 1, l1);
        l2 = max(res[0][i - 1] + 1, l2);
      }
      int r1 = p2;
      if (i + 1 < res[0].size()) {
        r1 = min(res[0][i + 1] - 1, r1);
        r2 = min(res[0][i + 1] - 1, r2);
      }
      if (r1 >= l1 && r2 >= l2) {
        ret += (r1 - l1 + 1.0) / (p2 - p1 + 1.0) * (r2 - l2 + 1.0) /
                   (v2 - v1 + 1.0) -
               max(0, min(res[0][i] - 1, r1) - l1 + 1) / (p2 - p1 + 1.0) *
                   max(0, min(res[0][i] - 1, r2) - l2 + 1) / (v2 - v1 + 1.0) -
               max(0, r1 - max(l1, res[0][i] + 1) + 1) / (p2 - p1 + 1.0) *
                   max(0, r2 - max(l2, res[0][i] + 1) + 1) / (v2 - v1 + 1.0);
      }
    }
    cout << setprecision(9) << ret << endl;
    return 0;
  }
  for (int i = 0; i < res[0].size(); i++) {
    if (i + k > res[0].size()) break;
    int l1 = min(res[0][i], p2);
    int r1 = max(res[0][i + k - 1], v1);
    int l2 = p1;
    int r2 = v2;
    if (i > 0) l2 = max(l2, res[0][i - 1] + 1);
    if (i + k < res[0].size()) r2 = min(r2, res[0][i + k] - 1);
    if (l1 >= l2 && r2 >= r1) {
      ret +=
          (l1 - l2 + 1.0) / (p2 - p1 + 1.0) * (r2 - r1 + 1.0) / (v2 - v1 + 1.0);
    }
  }
  sort(res[0].begin(), res[0].end(), greater<int>());
  for (int i = 0; i < res[0].size(); i++) {
    if (i + k > res[0].size()) break;
    int l1 = max(res[0][i], p1);
    int r1 = min(res[0][i + k - 1], v2);
    int l2 = p2;
    int r2 = v1;
    if (i > 0) l2 = min(l2, res[0][i - 1] - 1);
    if (i + k < res[0].size()) r2 = max(r2, res[0][i + k] + 1);
    if (l2 >= l1 && r2 <= r1)
      ret +=
          (l2 - l1 + 1.0) / (p2 - p1 + 1.0) * (r1 - r2 + 1.0) / (v2 - v1 + 1.0);
  }
  cout << setprecision(9) << ret << endl;
  return 0;
}