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;
vector<long long> s;
void rec(long long x) {
  if (x > 1e11) return;
  s.push_back(x);
  rec(x * 10 + 4);
  rec(x * 10 + 7);
}
long long f(long long l1, long long r1, long long l2, long long r2) {
  l1 = max(l1, l2);
  r1 = min(r1, r2);
  return max(r1 - l1 + 1, 0ll);
}
int main() {
  ios_base::sync_with_stdio(0);
  long long pl, printf, vl, vr, k;
  rec(0);
  sort(s.begin(), s.end());
  cin >> pl >> printf >> vl >> vr >> k;
  long long ans = 0;
  for (int i = 1; i + k < s.size(); i++) {
    long long l1 = s[i - 1] + 1, r1 = s[i], l2 = s[i + k - 1],
              r2 = s[i + k] - 1;
    long long a = f(l1, r1, vl, vr) * f(l2, r2, pl, printf);
    long long b = f(l1, r1, pl, printf) * f(l2, r2, vl, vr);
    ans += a + b;
    if (k == 1 && a > 0 && b > 0) ans--;
  }
  long long all = (printf - pl + 1) * (vr - vl + 1);
  cout << fixed << setprecision(12) << 1.0 * ans / all << endl;
}