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;
int countBits(int mask) {
  int res = 0;
  while (mask) mask &= (mask - 1), ++res;
  return res;
}
string toString(long long n) {
  stringstream ss;
  ss << n;
  return ss.str();
}
long long toNumber(string s) {
  stringstream ss;
  long long n;
  ss << s;
  ss >> n;
  return n;
}
vector<int> all;
void generate(long long curr) {
  if (curr > 1e9) return;
  all.push_back(curr);
  generate(curr * 10 + 4);
  generate(curr * 10 + 7);
}
long double getCommon(long double a, long double b, long double i,
                      long double j) {
  if (a < i && b < i) return 0;
  if (a > j && b > j) return 0;
  int start = max(a, i);
  int end = min(b, j);
  return end - start + 1;
}
int main() {
  int p1, p2, v1, v2, k;
  long double res = 0;
  generate(4), generate(7);
  sort(all.begin(), all.end());
  cin >> p1 >> p2 >> v1 >> v2 >> k;
  for (int start = (int)(0), _m = (int)(all.size()); start < _m; ++start) {
    int end = start + k - 1;
    if (end >= all.size()) break;
    long double pre = start == 0 ? 0 : all[start - 1] + 1;
    long double post = end == all.size() - 1 ? 1e9 : all[end + 1] - 1;
    long double a = getCommon(p1, p2, pre, all[start]);
    long double b = getCommon(v1, v2, all[end], post);
    res += a * b;
    a = getCommon(v1, v2, pre, all[start]);
    b = getCommon(p1, p2, all[end], post);
    res += a * b;
    if (k == 1)
      if (all[start] >= p1 && all[start] <= p2 && all[start] >= v1 &&
          all[start] <= v2)
        res--;
  }
  printf("%.12f\n", res / (p2 - p1 + 1) / (v2 - v1 + 1));
}