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 long long mod = 1e9 + 7;
const long long INF = 2e9;
const long long N = 1e5 + 15;
vector<string> split(const string& s, char c) {
  vector<string> v;
  stringstream ss(s);
  string x;
  while (getline(ss, x, c)) v.emplace_back(x);
  return move(v);
}
template <typename T, typename... Args>
inline string arrStr(T arr, long long n) {
  stringstream s;
  s << "[";
  for (long long i = 0; i < n - 1; i++) s << arr[i] << ",";
  s << arr[n - 1] << "]";
  return s.str();
}
inline void __evars_begin(long long line) { cerr << "#" << line << ": "; }
template <typename T>
inline void __evars_out_var(vector<T> val) {
  cerr << arrStr(val, val.size());
}
template <typename T>
inline void __evars_out_var(T* val) {
  cerr << arrStr(val, 10);
}
template <typename T>
inline void __evars_out_var(T val) {
  cerr << val;
}
inline void __evars(vector<string>::iterator it) { cerr << endl; }
template <typename T, typename... Args>
inline void __evars(vector<string>::iterator it, T a, Args... args) {
  cerr << it->substr((*it)[0] == ' ', it->length()) << "=";
  __evars_out_var(a);
  cerr << "; ";
  __evars(++it, args...);
}
long long ovl(long long x, long long y, long long l, long long r) {
  long long mx = max(x, l), mn = min(y, r);
  return max(0LL, mn - mx + 1);
}
void kudne() {
  vector<long long> gen(1);
  for (long long i = 1; i <= 9; ++i) {
    for (long long j = 0; j < (1 << i); ++j) {
      long long t = 0;
      for (long long k = 0; k < i; ++k) {
        if (j & (1 << k))
          t = t * 10 + 7;
        else
          t = t * 10 + 4;
      }
      gen.push_back(t);
    }
  }
  long long sz = gen.size();
  gen.push_back(1e9 + 1);
  long long l1, r1, l2, r2, k;
  cin >> l1 >> r1 >> l2 >> r2 >> k;
  long long ans = 0;
  sort(gen.begin(), gen.end());
  for (long long i = 1; i + k <= sz; ++i) {
    ans += ovl(l1, r1, gen[i - 1] + 1, gen[i]) *
           ovl(l2, r2, gen[i + k - 1], gen[i + k] - 1);
    ans += ovl(l2, r2, gen[i - 1] + 1, gen[i]) *
           ovl(l1, r1, gen[i + k - 1], gen[i + k] - 1);
    if (k == 1) {
      if (l1 <= gen[i] && gen[i] <= r1 && l2 <= gen[i] && gen[i] <= r2) ans--;
    }
  }
  long double d = (r1 - l1 + 1) * (r2 - l2 + 1), n = ans;
  n *= 1.0;
  d *= 1.0;
  n = n / d;
  cout << setprecision(10) << n;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long tc = 1;
  for (long long i = 1; i <= tc; ++i) {
    kudne();
    if (i < tc) cout << "\n";
  }
  return 0;
}