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;
ofstream file("debug.txt");
template <typename Int>
ostream &operator<<(ostream &s, const vector<Int> &v) {
  s << '[';
  for (auto i = v.begin(); i != v.end(); i++) {
    s << *i;
    if (i != prev(v.end())) s << ", ";
  }
  s << ']';
  return s;
}
template <typename Int, typename INT>
ostream &operator<<(ostream &s, const map<Int, INT> &mp) {
  s << '[';
  for (auto &i : mp) {
    s << '[' << i.first << ", " << i.second << ']';
    if (i != *mp.rbegin()) s << ", ";
  }
  s << ']';
  return s;
}
template <typename Int>
ostream &operator<<(ostream &s, const set<Int> &st) {
  s << '[';
  for (auto &i : st) {
    s << i;
    if (i != *st.rbegin()) s << ", ";
  }
  s << ']';
  s << '\n';
  return s;
}
template <typename Int, typename INt>
ostream &operator<<(ostream &s, const pair<Int, INt> &x) {
  s << '[' << x.first << ", " << x.second << ']';
  return s;
}
template <typename T>
void print(T *x, int n) {
  for (int i = 0; i < n; i++) {
    file << x[i] << " \n"[i == n - 1];
  }
}
template <typename T>
ostream &operator<<(ostream &s, const complex<T> &a) {
  s << '(' << a.real() << ", " << a.imag() << ')';
  return s;
}
const int N = 1e5 + 5;
long long pl, pr, vl, vr, k, x[1026];
long long next(long long x) {
  long long y = x, tmp = 1;
  while (1) {
    if (x % 10 == 0) {
      y += 4 * tmp;
      break;
    }
    if (x % 10 == 4) {
      y += 3 * tmp;
      break;
    }
    y -= 3 * tmp;
    x /= 10;
    tmp *= 10;
  }
  return y;
}
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  for (int i = 1; i < 1026; i++) x[i] = next(x[i - 1]);
  cin >> pl >> pr >> vl >> vr >> k;
  int l = 1, r = k;
  long long ans = 0, mnl, mxl, mnr, mxr;
  while (r < 1026) {
    mnl = max(x[l - 1], pl - 1);
    mxl = min(x[l], pr);
    mnr = max(x[r], vl);
    mxr = min(x[r + 1], vr + 1);
    if (mxr >= mnr && mxl >= mnl) ans += (mxl - mnl) * (mxr - mnr);
    mnl = max(x[l - 1], vl - 1);
    mxl = min(x[l], vr);
    mnr = max(x[r], pl);
    mxr = min(x[r + 1], pr + 1);
    if (mxr >= mnr && mxl >= mnl) ans += (mxl - mnl) * (mxr - mnr);
    if (k == 1 && pl <= x[l] && x[l] <= pr && vl <= x[l] && x[l] <= vr) ans--;
    l++;
    r++;
  }
  cout << fixed << setprecision(12)
       << 1.0 * ans / (pr - pl + 1) / (vr - vl + 1);
}