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;
double getnum(long int l, long int r, long int a, long int b) {
  l = max(l, a);
  r = min(r, b);
  if (l > r) return 0;
  return (double)(r - l + 1) / (b - a + 1);
}
int main() {
  int T;
  long int Pl, Pr, Vl, Vr, k, i, l1, r1, l2, r2, n;
  double ans;
  vector<long int> a;
  a.push_back(0);
  for (n = 1; n <= 9; ++n) {
    for (long int I = 0; I < (1 << n); ++I) {
      long int x = 0;
      for (int i = 0; i < n; ++i) {
        if (I & (1 << i))
          x = x * 10 + 4;
        else
          x = x * 10 + 7;
      }
      a.push_back(x);
    }
  }
  a.push_back(1000000001);
  sort(a.begin(), a.end());
  n = a.size();
  cin >> Pl >> Pr >> Vl >> Vr >> k;
  ans = 0;
  for (i = 1; i + k < n + 1; ++i) {
    l1 = a[i - 1] + 1;
    r1 = a[i];
    l2 = a[i + k - 1];
    r2 = a[i + k] - 1;
    ans += getnum(l1, r1, Pl, Pr) * getnum(l2, r2, Vl, Vr);
    ans += getnum(l2, r2, Pl, Pr) * getnum(l1, r1, Vl, Vr);
    if (r1 == l2) ans -= getnum(r1, l2, Pl, Pr) * getnum(r1, l2, Vl, Vr);
  }
  cout << setprecision(10) << fixed << ans << setprecision(0) << endl;
  return 0;
}