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>
#pragma GCC optimize("O3")
using namespace std;
vector<long long> vec;
long long pl, pr, vl, vr, k;
long double p(long long a, long long b, long long c, long long d) {
  long long l = max(a, c);
  long long r = min(b, d);
  long double x = max(0LL, r - l + 1);
  x /= (long double)(b - a + 1);
  return x;
}
long double f(long long l, long long r, long long a, long long b, long long c,
              long long d) {
  return (long double)p(a, b, (l == 0 ? 0 : vec[l - 1] + 1), vec[l]) *
         p(c, d, vec[r],
           (r == (long long)((vec).size()) - 1 ? 1e9 : vec[r + 1] - 1));
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cerr.tie(0);
  cout.tie(0);
  setprecision(100);
  for (long long x = 1; x <= 9; x++) {
    for (long long mask = 0; mask < (1 << x); mask++) {
      string s = "";
      for (long long j = 0; j < x; j++) {
        if (mask & (1 << j))
          s += '4';
        else
          s += '7';
      }
      vec.push_back(stoi(s));
    }
  }
  sort(vec.begin(), vec.end());
  cin >> pl >> pr >> vl >> vr >> k;
  long double ans = 0;
  for (long long i = 0; i + k - 1 < (long long)((vec).size()); i++) {
    ans += f(i, i + k - 1, pl, pr, vl, vr);
    ans += f(i, i + k - 1, vl, vr, pl, pr);
    if (k == 1 && (f(i, i + k - 1, pl, pr, vl, vr)) &&
        f(i, i + k - 1, vl, vr, pl, pr))
      ans -= (long double)(1) / ((vr - vl + 1) * (pr - pl + 1));
  }
  return cout << fixed << setprecision(20) << ans, 0;
}