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>
inline string tostr(const _T& a) {
  ostringstream os("");
  os << a;
  return os.str();
}
long long ar, br, al, bl, k;
vector<long long> v;
void rec1(long long x) {
  if (x > br && x > ar) return;
  if (x) v.push_back(x);
  rec1(x * 10 + 4);
  rec1(x * 10 + 7);
}
long double prob(long long L, long long R, long long LL, long long RR) {
  if (R < LL || L > RR) return 0;
  if (L < LL) L = LL;
  if (R > RR) R = RR;
  return ((R - L + 1) * 1.0) / ((RR - LL + 1) * 1.0);
}
int main() {
  cout.flags(ios::fixed);
  cout.precision(15);
  cin >> al >> ar >> bl >> br >> k;
  v.push_back(0);
  rec1(0);
  v.push_back(max(ar, br) + 1);
  long double p = 0;
  sort(v.begin(), v.end());
  for (int i = 1; i < v.size(); i++) {
    if (i + k >= v.size()) {
      break;
    }
    p += prob(v[i - 1] + 1, v[i], al, ar) *
         prob(v[i + k - 1], v[i + k] - 1, bl, br);
    p += prob(v[i - 1] + 1, v[i], bl, br) *
         prob(v[i + k - 1], v[i + k] - 1, al, ar);
    if (k == 1) {
      p -= prob(v[i], v[i], al, ar) * prob(v[i + k - 1], v[i + k - 1], bl, br);
    }
  }
  cout << p;
  return 0;
}