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;
struct point {
  double x, y;
  point(double x, double y) : x(x), y(y) {}
  point() { x = y = 0.0; }
  point operator+(point q) { return point(x + q.x, y + q.y); }
  point operator-(point q) { return point(x - q.x, y - q.y); }
  point operator*(double t) { return point(x * t, y * t); }
  point operator/(double t) { return point(x / t, y / t); }
  double operator*(point q) { return q.x * x + q.y * y; }
  double operator%(point q) { return x * q.y - y * q.x; }
};
const double PI = 2.0 * acos(0.0);
inline int getInt() {
  int x;
  scanf("%d", &x);
  return x;
}
inline long long getLL() {
  long long x;
  scanf("%lld", &x);
  return x;
};
inline int NUM(char c) { return (int)c - 48; }
inline char CHR(int n) { return (char)(n + 48); }
template <class T>
inline T MAX(T a, T b) {
  if (a > b) return a;
  return b;
}
template <class T>
inline T MIN(T a, T b) {
  if (a < b) return a;
  return b;
}
inline int BPM(int a, int b, int MOD) {
  if (a == 0) return 0;
  if (b == 0) return 1 % MOD;
  if (b % 2 == 1) return (int)((long long)BPM(a, b - 1, MOD) * a % MOD);
  int tmp = BPM(a, b / 2, MOD);
  return (int)((long long)tmp * tmp % MOD);
}
inline int phi(int N) {
  int ret = N;
  for (int i = 2; N > 1; i++) {
    if (i * i > N) {
      ret -= ret / N;
      break;
    }
    if (N % i == 0) {
      ret -= ret / i;
      while (N % i == 0) N /= i;
    }
  }
  return ret;
}
inline void OPEN(string s) {
  freopen((s + ".in").c_str(), "r", stdin);
  freopen((s + ".out").c_str(), "w", stdout);
}
vector<int> v;
inline void generate_all_lucky() {
  for (int p = (1), _b = (9); p <= _b; p++) {
    for (int bm = 0, _n = (1 << p); bm < _n; bm++) {
      int tmp = 0;
      for (int i = 0, _n = (p); i < _n; i++) {
        if (bm & (1 << i)) {
          tmp = tmp * 10 + 4;
        } else {
          tmp = tmp * 10 + 7;
        }
      }
      v.push_back(tmp);
    }
  }
}
int main() {
  int pl, pr, ql, qr;
  scanf("%d%d%d%d", &pl, &pr, &ql, &qr);
  int k;
  scanf("%d", &k);
  generate_all_lucky();
  sort((v).begin(), (v).end());
  int vsz = (v).size();
  double ret = 0.0;
  for (int i = 0, _n = ((v).size()); i < _n; i++) {
    if (i + k - 1 <= (v).size() - 1) {
      int a, b, c, d;
      int _p1, _p2, _q1, _q2;
      _p1 = pl;
      _p2 = pr;
      _q1 = ql;
      _q2 = qr;
      if (i - 1 < 0)
        a = 1;
      else
        a = v[i - 1] + 1;
      a = max(a, _p1);
      b = v[i];
      b = min(b, _p2);
      c = v[i + k - 1];
      c = max(c, _q1);
      if (i + k >= vsz)
        d = 1000000000;
      else
        d = v[i + k] - 1;
      d = min(d, _q2);
      if (a <= b && c <= d) {
        ret += 1. * (d - c + 1) * (b - a + 1);
        if (b == c) ret -= 1.0;
      }
      _p1 = ql;
      _p2 = qr;
      _q1 = pl;
      _q2 = pr;
      if (i - 1 < 0)
        a = 1;
      else
        a = v[i - 1] + 1;
      a = max(a, _p1);
      b = v[i];
      b = min(b, _p2);
      c = v[i + k - 1];
      c = max(c, _q1);
      if (i + k >= vsz)
        d = 1000000000;
      else
        d = v[i + k] - 1;
      d = min(d, _q2);
      if (a <= b && c <= d) {
        ret += 1. * (d - c + 1) * (b - a + 1);
      }
    } else
      break;
  }
  ret /= (qr - ql + 1);
  ret /= (pr - pl + 1);
  printf("%.12lf\n", ret);
  return 0;
}