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>
template <typename T>
T inline SQR(const T& a) {
  return a * a;
}
template <typename T>
T inline ABS(const T& a) {
  return a < 0 ? -a : a;
}
template <typename T>
T inline MIN(const T& a, const T& b) {
  if (a < b) return a;
  return b;
}
template <typename T>
T inline MAX(const T& a, const T& b) {
  if (a > b) return a;
  return b;
}
const double EPS = 1e-9;
inline int SGN(double a) { return a > EPS ? 1 : (a < -EPS ? -1 : 0); }
inline int CMP(double a, double b) { return SGN(a - b); }
using namespace std;
vector<long long> lucky;
void build(long long now) {
  if (now > 1000000000LL) return;
  lucky.push_back(10LL * now + 4LL);
  build(10LL * now + 4LL);
  lucky.push_back(10LL * now + 7LL);
  build(10LL * now + 7LL);
}
double prob(double a, double b, double c, double d) {
  return (min(d, b) - max(c, a) + 1) / (d - c + 1);
}
int main(int argc, char* argv[]) {
  ios::sync_with_stdio(false);
  build(0);
  sort(lucky.begin(), lucky.end());
  long long pl, pr, vl, vr, k;
  cin >> pl >> pr >> vl >> vr >> k;
  double ans = 0.0;
  for (int i = 0; i + k - 1 < int((lucky).size()); i++) {
    long long lo = lucky[i];
    long long hi = lucky[i + k - 1];
    long long alo, ahi;
    if (i)
      alo = lucky[i - 1] + 1;
    else
      alo = 1;
    if (i + k < int((lucky).size()))
      ahi = lucky[i + k] - 1;
    else
      ahi = 1000000000;
    double p1, p2;
    p1 = 0.0;
    p2 = 0.0;
    if ((pl <= alo && alo <= pr) || (alo <= pl && pl <= lo))
      p1 = prob(alo, lo, pl, pr);
    if ((vl <= hi && hi <= vr) || (hi <= vl && vl <= ahi))
      p2 = prob(hi, ahi, vl, vr);
    ans += p1 * p2;
    p1 = 0.0;
    p2 = 0.0;
    if ((vl <= alo && alo <= vr) || (alo <= vl && vl <= lo))
      p1 = prob(alo, lo, vl, vr);
    if ((pl <= hi && hi <= pr) || (hi <= pl && pl <= ahi))
      p2 = prob(hi, ahi, pl, pr);
    ans += p1 * p2;
    if (k == 1 && pl <= lucky[i] && lucky[i] <= pr && vl <= lucky[i] &&
        lucky[i] <= vr)
      ans -=
          prob(lucky[i], lucky[i], pl, pr) * prob(lucky[i], lucky[i], vl, vr);
  }
  cout << fixed << setprecision(20) << ans << "\n";
  return 0;
}