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;
template <class T>
bool checkmax(T& a, T b) {
  return a < b ? a = b, 1 : 0;
}
template <class T>
bool checkmin(T& a, T b) {
  return a > b ? a = b, 1 : 0;
}
template <class T>
void string_reader(string s, vector<T>& vec) {
  istringstream sin(s);
  copy(istream_iterator<T>(sin), istream_iterator<T>(), back_inserter(vec));
}
vector<long long> luck;
const int N = 1000000001;
void build(long long s) {
  if (s >= N) return;
  luck.push_back(s);
  build(s * 10 + 4);
  build(s * 10 + 7);
}
double radio(double l, double r, double pl, double pr) {
  return max(0.0, (r - l + 1) / (pr - pl + 1));
}
double get_res(double pl, double pr, double vl, double vr, int k,
               vector<tuple<double, double, double, double> >& hash) {
  double res = 0;
  int n = luck.size();
  for (int i = 0; i < n - k - 1; i++) {
    double bl = luck[i] + 1, br = luck[i + 1];
    double el = luck[i + k], er = luck[i + k + 1] - 1;
    double vbl = max(bl, pl), vbr = min(br, pr);
    double vel = max(el, vl), ver = min(er, vr);
    double add = radio(vbl, vbr, pl, pr) * radio(vel, ver, vl, vr);
    res += add;
    if (add != 0) {
      hash.push_back(make_tuple(vbl, vbr, vel, ver));
    }
  }
  return res;
}
int main() {
  build(0);
  luck.push_back(N);
  sort(luck.begin(), luck.end());
  int pl, pr, vl, vr, k;
  while (scanf("%d %d %d %d %d", &pl, &pr, &vl, &vr, &k) == 5) {
    vector<tuple<double, double, double, double> > p, l;
    double res = 0;
    res += get_res(pl, pr, vl, vr, k, p);
    res += get_res(vl, vr, pl, pr, k, l);
    for (const tuple<double, double, double, double>& i : p) {
      for (const tuple<double, double, double, double>& j : l) {
        double pll = max(get<2>(j), get<0>(i)), prr = min(get<3>(j), get<1>(i));
        double vll = max(get<2>(i), get<0>(j)), vrr = min(get<3>(i), get<1>(j));
        res -= radio(pll, prr, pl, pr) * radio(vll, vrr, vl, vr);
      }
    }
    printf("%.10f\n", res);
  }
  return 0;
}