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 T MIN(T a, T b) {
  return a < b ? a : b;
}
template <class T>
inline T MAX(T a, T b) {
  return a > b ? a : b;
}
template <class T>
inline void getInt(T& x) {
  char c;
  int mul = 1;
  while ((c = getchar()) != EOF) {
    if (c == '-') mul = -1;
    if (c >= '0' && c <= '9') {
      x = c - '0';
      break;
    }
  }
  if (c == EOF) {
    x = EOF;
    return;
  }
  while (c = getchar()) {
    if (c >= '0' && c <= '9') {
      x = (x << 1) + (x << 3) + (c - '0');
    } else
      break;
  }
  x *= mul;
}
long long k, a, b, c, d;
vector<long long> v;
void gen(string s, int l) {
  if (s.length() == l) {
    if (atoll(s.c_str()) < 0) cout << s << endl;
    v.push_back(atoll(s.c_str()));
  } else {
    gen(s + "4", l);
    gen(s + "7", l);
  }
}
int main() {
  v.push_back(0);
  for (int i = (1), _b = (10); i <= _b; i++) {
    gen("", i);
  }
  sort(v.begin(), v.end());
  getInt(a);
  getInt(b);
  getInt(c);
  getInt(d);
  getInt(k);
  double ok = 0;
  for (int i = (1), _b = (v.size() - 1); i <= _b; i++) {
    int j = i + k - 1;
    if (j < v.size()) {
      if (!(a > v[i] || b <= v[i - 1] || c >= v[j + 1] || d < v[j])) {
        long long lo = MIN(v[i], b) - MAX(v[i - 1] + 1, a) + 1;
        long long hi = MIN(v[j + 1] - 1, d) - MAX(v[j], c) + 1;
        if (lo > 0 && hi > 0) {
          ok += 1.0 * lo / (b - a + 1) * hi / (d - c + 1);
        }
      }
      if (!(c > v[i] || d <= v[i - 1] || a >= v[j + 1] || b < v[j])) {
        long long lo = MIN(v[i], d) - MAX(v[i - 1] + 1, c) + 1;
        long long hi = MIN(v[j + 1] - 1, b) - MAX(v[j], a) + 1;
        if (lo > 0 && hi > 0) {
          ok += 1.0 * lo / (d - c + 1) * hi / (b - a + 1);
        }
      }
      if (k == 1 && a <= v[i] && b >= v[i] && c <= v[j] && d >= v[j]) {
        ok -= 1.0 / (d - c + 1) / (b - a + 1);
      }
    }
  }
  printf("%.12f\n", ok);
  return 0;
}