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<long long> v;
int main() {
  long long pl, pr, vl, vr, k;
  double ans = 0;
  cin >> pl >> pr >> vl >> vr >> k;
  for (int h = 1; h <= 9; h++) {
    for (int i = 0; i < (1 << h); i++) {
      int x = 0, y = 1;
      for (int j = 0; j < h; j++) {
        if (i & (1 << j)) {
          x += 4 * y;
        } else {
          x += 7 * y;
        }
        y *= 10;
      }
      v.push_back(x);
    }
  }
  v.push_back(0);
  sort(v.begin(), v.end());
  v.push_back(1e9 + 1);
  double pt = 0;
  for (int i = 0; i + k + 1 < v.size(); i++) {
    ans += max(0ll, (min(v[i + 1], pr) - max(v[i] + 1, pl)) + 1) *
           max(0ll, (min(v[i + k + 1] - 1, vr) - max(v[i + k], vl)) + 1);
    ans += max(0ll, (min(v[i + 1], vr) - max(v[i] + 1, vl)) + 1) *
           max(0ll, (min(v[i + k + 1] - 1, pr) - max(v[i + k], pl)) + 1);
    if (pt != ans && k == 1) {
      if (min(v[i + 1], pr) == max(v[i + k], vl) ||
          max(v[i] + 1, vl) == min(v[i + k + 1] - 1, pr)) {
        ans--;
      }
    }
    pt = ans;
  }
  double all = ((pr - pl) + 1) * ((vr - vl) + 1);
  cout << fixed << setprecision(12) << ans / all;
  return 0;
}