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;
using vi = vector<int>;
using vvi = vector<vi>;
using ii = pair<int, int>;
using vii = vector<ii>;
using l = long long;
using vl = vector<l>;
using vvl = vector<vl>;
using ll = pair<l, l>;
using vll = vector<ll>;
using vvll = vector<vll>;
using lu = unsigned long long;
using vb = vector<bool>;
using vvb = vector<vb>;
using vd = vector<double>;
using vvd = vector<vd>;
const int INF = numeric_limits<int>::max();
const double EPS = 1e-10;
const l e5 = 100000, e6 = 1000000, e7 = 10000000, e9 = 1000000000;
const bool enable_log = false;
struct VoidStream {
  void operator&(std::ostream&) {}
};
l intersect(l a, l b, l c, l d) { return max((l)0, min(b, d) - max(a, c) + 1); }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  vl lucky = {0, 4, 7};
  {
    l from = 1;
    for (l i = 0; i < 9; i++) {
      l n = lucky.size();
      for (l j = from; j < n; j++) {
        lucky.emplace_back(lucky[j] * 10 + 4);
        lucky.emplace_back(lucky[j] * 10 + 7);
      }
      from = n;
    }
  }
  l pl, pr, ul, ur, k;
  while (cin >> pl >> pr >> ul >> ur >> k) {
    auto a = lucky.begin();
    auto b = ((k + 1) < lucky.size()) ? (lucky.begin() + k + 1) : lucky.end();
    l total = 0;
    while (b != lucky.end()) {
      l al = *a + 1, ar = *(a + 1), bl = *(b - 1), br = *b - 1;
      l d = intersect(al, ar, pl, pr) * intersect(bl, br, ul, ur) +
            intersect(al, ar, ul, ur) * intersect(bl, br, pl, pr);
      if (k == 1 and intersect(pl, pr, ar, ar) and intersect(ul, ur, ar, ar))
        d--;
      total += d;
      a++;
      b++;
    }
    cout.precision(12);
    cout << fixed << ((double)total) / ((double)((pr - pl + 1) * (ur - ul + 1)))
         << '\n';
  }
}