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;
long long f(int a, int b, int l, int r) {
  return max(0, min(b, r) - max(a, l) + 1);
}
int main() {
  int a, b, l, r, k;
  cin >> a >> b >> l >> r >> k;
  vector<int> n;
  for (int I = 0; I <= 9; ++I)
    for (int j = 0; j < 1 << I; ++j) {
      int v{};
      for (int i = 0, t{1}; i < I; ++i, t *= 10)
        if (j >> i & 1)
          v += 7 * t;
        else
          v += 4 * t;
      n.push_back(v);
    }
  n.push_back(1000000000);
  long long c{};
  for (size_t i = 1; i < n.size() - k; ++i) {
    c += f(n[i - 1] + 1, n[i], a, b) * f(n[i + k - 1], n[i + k] - 1, l, r) +
         f(n[i - 1] + 1, n[i], l, r) * f(n[i + k - 1], n[i + k] - 1, a, b);
    if (k == 1 && f(n[i], n[i], a, b) * f(n[i], n[i], l, r) == 1) --c;
  }
  cout << fixed << setprecision(9)
       << (long double)c / (b - a + 1) / (r - l + 1);
}