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 di[] = {0, 0, 1, -1, 1, 1, -1, -1};
int dj[] = {1, -1, 0, 0, 1, -1, 1, -1};
vector<int> luckyNumbers;
vector<pair<int, int> > p, v;
void init() {
  int lvl = 0, pos = 0;
  luckyNumbers.push_back(0);
  while (lvl < 9) {
    int lim = ((int)luckyNumbers.size());
    while (pos < lim) {
      int st = luckyNumbers[pos++];
      luckyNumbers.push_back((st * 10) + 4);
      luckyNumbers.push_back((st * 10) + 7);
    }
    ++lvl;
  }
  luckyNumbers.push_back(1000000001);
}
void makePairs(int frm, int to, vector<pair<int, int> > &vec) {
  for (int i = 0; i < ((int)luckyNumbers.size()); ++i) {
    if (frm > luckyNumbers[i]) continue;
    if (frm > to) break;
    vec.push_back(make_pair(frm, min(to, max(frm, luckyNumbers[i] - 1))));
    if (frm < luckyNumbers[i] && luckyNumbers[i] < to)
      vec.push_back(make_pair(luckyNumbers[i], luckyNumbers[i]));
    frm = vec.back().second + 1;
  }
}
int countLucky(int frm, int to) {
  return upper_bound(luckyNumbers.begin(), luckyNumbers.end(), to) -
         lower_bound(luckyNumbers.begin(), luckyNumbers.end(), frm);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout << fixed << setprecision(10);
  init();
  int p1, p2, v1, v2, k;
  cin >> p1 >> p2 >> v1 >> v2 >> k;
  makePairs(p1, p2, p);
  makePairs(v1, v2, v);
  double lucky = 0.0;
  for (int i = 0; i < ((int)p.size()); ++i) {
    for (int j = 0; j < ((int)v.size()); ++j) {
      int szp = p2 - p1 + 1, szv = v2 - v1 + 1;
      pair<int, int> a = p[i], b = v[j];
      if (a > b) swap(a, b), swap(szp, szv);
      if (countLucky(a.second, b.first) == k) {
        lucky += ((1.0 * (a.second - a.first + 1)) / szp) *
                 ((1.0 * (b.second - b.first + 1)) / szv);
      }
    }
  }
  cout << lucky << '\n';
  return 0;
}