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;
const int oo = (int)1e9, MAXN = (int)1e6, risk = (int)1e6;
const double eps = 1e-9;
vector<int> lucky;
void find(long long l) {
  if (l >= 1000000000) return;
  lucky.push_back(l);
  find(l * 10 + 4);
  find(l * 10 + 7);
}
vector<pair<pair<int, int>, pair<int, int> > > quad;
void findquad(int k) {
  int c = 0, lo = 0;
  for (int i = (k - 1); i <= (lucky.size() - 1); i++) {
    pair<int, int> l = make_pair(lo + 1, lucky[i - k + 1]);
    pair<int, int> r = make_pair(
        lucky[i], (i == lucky.size() - 1) ? 1000000000 : lucky[i + 1] - 1);
    quad.push_back(make_pair(l, r));
    lo = lucky[i - k + 1];
  }
}
long long intersect(long long x1, long long x2, long long y1, long long y2) {
  if (x1 > y1) swap(x1, y1), swap(x2, y2);
  if (x2 < y1) return 0;
  if (x2 > y2) return y2 - y1 + 1;
  return x2 - y1 + 1;
}
char w[] = {'a', 'b', 'c', 'd'};
int main() {
  long long a, a1, a2, b, b1, b2, k, l, r;
  find(4);
  find(7);
  sort((lucky).begin(), (lucky).end());
  cin >> a1 >> a2 >> b1 >> b2 >> k;
  a = a2 - a1 + 1, b = b2 - b1 + 1;
  findquad(k);
  double ans = 0;
  for (int i = 0; i < (quad.size()); i++) {
    long long l1 = quad[i].first.first;
    long long l2 = quad[i].first.second;
    long long r1 = quad[i].second.first;
    long long r2 = quad[i].second.second;
    l = intersect(l1, l2, a1, a2);
    r = intersect(r1, r2, b1, b2);
    ans += double(l * r) / double(a * b);
    l = intersect(l1, l2, b1, b2);
    r = intersect(r1, r2, a1, a2);
    ans += double(l * r) / double(a * b);
  }
  if (k == 1) {
    for (int i = 0; i < (lucky.size()); i++)
      if (a1 <= lucky[i] && a2 >= lucky[i] && b1 <= lucky[i] && b2 >= lucky[i])
        ans -= 1.0 / double(a * b);
  }
  printf("%.10lf\n", ans);
  return 0;
}