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 int M = 1000 + 10;
int main() {
  int pl, pr, vl, vr, k;
  cin >> pl >> pr >> vl >> vr >> k;
  vector<int> candidate;
  for (int d = 1; d <= 9; d++) {
    for (int i = 0; i < (1 << d); i++) {
      int num = 0;
      for (int r = 0; r < d; r++) {
        if ((i >> r) & 1) {
          num = num * 10 + 7;
        } else {
          num = num * 10 + 4;
        }
      }
      candidate.push_back(num);
    }
  }
  sort(candidate.begin(), candidate.end());
  long long int ans = 0;
  for (int i = 0; i < candidate.size(); i++) {
    int leftborder1 = i >= 1 ? candidate[i - 1] + 1 : 1;
    int rightborder1 = candidate[i];
    int leftborder2 = -1;
    int rightborder2 = -1;
    if (i + k - 1 < candidate.size()) {
      leftborder2 = candidate[i + k - 1];
      rightborder2 = i + k < candidate.size() ? candidate[i + k] - 1 : 1E+9;
      long long int num1left =
          min(pr, rightborder1) - max(pl, leftborder1) >= 0
              ? min(pr, rightborder1) - max(pl, leftborder1) + 1
              : 0;
      long long int num2right =
          min(vr, rightborder2) - max(vl, leftborder2) >= 0
              ? min(vr, rightborder2) - max(vl, leftborder2) + 1
              : 0;
      ans = ans + num1left * num2right;
      long long int num1right =
          min(pr, rightborder2) - max(pl, leftborder2) >= 0
              ? min(pr, rightborder2) - max(pl, leftborder2) + 1
              : 0;
      long long int num2left =
          min(vr, rightborder1) - max(vl, leftborder1) >= 0
              ? min(vr, rightborder1) - max(vl, leftborder1) + 1
              : 0;
      ans = ans + num1right * num2left;
      if (k == 1) {
        if (min(pr, candidate[i]) - max(pl, candidate[i]) >= 0 &&
            min(vr, candidate[i]) - max(vl, candidate[i]) >= 0) {
          ans--;
        }
      }
    }
  }
  printf("%.15f\n", 1.0 * ans / (pr - pl + 1) / (vr - vl + 1));
  return 0;
}