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;
vector<int> v;
int A1, A2, B1, B2, k;
long long f(long long a, long long b, long long c, long long d) {
  if (max(a, c) <= min(b, d)) return 1 + min(b, d) - max(a, c);
  return 0;
}
int main() {
  string s;
  int tmp, KK;
  for (int i = 0; i <= 9; i++)
    for (int j = 0; j + i <= 9; j++) {
      s = "";
      if (i + j == 0) continue;
      for (int z = 0; z < i; z++) s += "4";
      for (int z = 0; z < j; z++) s += "7";
      do {
        stringstream ss;
        ss << s;
        ss >> tmp;
        v.push_back(tmp);
      } while (next_permutation(s.begin(), s.end()));
    }
  sort(v.begin(), v.end());
  cin >> A1 >> A2 >> B1 >> B2 >> k;
  KK = k;
  int s1 = 0, s2 = 4, e1, e2;
  if (k > v.size()) {
    printf("%.11lf\n", 0);
    return 0;
  }
  v.push_back(1000000001);
  long long rt = 0;
  int cnt = 0;
  for (; k < v.size(); k++) {
    e1 = v[k - 1];
    e2 = v[k] - 1;
    rt += f(s1, s2, A1, A2) * f(e1, e2, B1, B2);
    rt += f(e1, e2, A1, A2) * f(s1, s2, B1, B2);
    s1 = v[cnt] + 1;
    s2 = v[cnt + 1];
    cnt++;
  }
  if (KK == 1)
    for (int i = 0; i < v.size(); i++)
      if (f(v[i], v[i], A1, A2) && f(v[i], v[i], B1, B2)) rt--;
  double ret = 0.0;
  ret = double(rt) / double(B2 - B1 + 1) / double(A2 - A1 + 1);
  printf("%.11lf\n", ret);
  return 0;
}