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;
pair<int, int> un(pair<int, int> a, pair<int, int> b) {
  if (a == make_pair(-1, -1) || b == make_pair(-1, -1))
    return make_pair(-1, -1);
  if (a > b) swap(a, b);
  if (a.second >= b.second)
    return b;
  else if (a.second < b.first)
    return make_pair(-1, -1);
  else
    return make_pair(b.first, a.second);
}
int sz(pair<int, int> a) {
  return a == make_pair(-1, -1) ? 0 : a.second - a.first + 1;
}
pair<int, int> p, v;
int n, k;
long long c1, c2;
vector<int> buf;
int main(void) {
  cin >> p.first >> p.second >> v.first >> v.second >> k;
  buf.push_back(0);
  for (int d = (1); d <= (9); ++d)
    for (int i = 0; i < ((1 << d)); ++i) {
      int tmp = 0;
      for (int j = 0; j < (d); ++j) {
        if (i & (1 << j))
          tmp = tmp * 10 + 7;
        else
          tmp = tmp * 10 + 4;
      }
      buf.push_back(tmp);
    }
  buf.push_back(1000000001);
  sort(buf.begin(), buf.end());
  n = buf.size() - 2;
  c2 = (long long)sz(p) * sz(v);
  for (int i = 1; i <= (n + 1 - k); ++i) {
    c1 += (long long)sz(un(make_pair(buf[i - 1] + 1, buf[i]), p)) *
          sz(un(make_pair(buf[i + k - 1], buf[i + k] - 1), v));
    c1 += (long long)sz(un(make_pair(buf[i - 1] + 1, buf[i]), v)) *
          sz(un(make_pair(buf[i + k - 1], buf[i + k] - 1), p));
    if (k == 1 && un(make_pair(buf[i], buf[i]), p) != make_pair(-1, -1) &&
        un(make_pair(buf[i], buf[i]), v) != make_pair(-1, -1))
      c1--;
  }
  if (c1 > c2) c1 = c2;
  printf("%.20lf\n", (double)c1 / c2);
  return 0;
}