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;
bool comp(string a, string b) {
  if (a.compare(b) < 0) return true;
  return false;
}
long long p1, p2, v1, v2, k;
long a[10000];
long long ans = 0;
long n = 1;
long maxlevel;
void res(long x1, long x2, long y1, long y2) {
  long long o1, o2, t1, t2;
  o1 = x1 > p1 ? x1 : p1;
  o2 = x2 < p2 ? x2 : p2;
  o1 = o1 <= o2 ? o2 - o1 + 1 : 0;
  t1 = y1 > v1 ? y1 : v1;
  t2 = y2 < v2 ? y2 : v2;
  t1 = t1 <= t2 ? t2 - t1 + 1 : 0;
  ans += t1 * o1;
  o1 = x1 > v1 ? x1 : v1;
  o2 = x2 < v2 ? x2 : v2;
  o1 = o1 <= o2 ? o2 - o1 + 1 : 0;
  t1 = y1 > p1 ? y1 : p1;
  t2 = y2 < p2 ? y2 : p2;
  t1 = t1 <= t2 ? t2 - t1 + 1 : 0;
  ans += t1 * o1;
  if ((x2 == y1) && (x2 >= p1) && (x2 <= p2)) ans -= 1;
}
void pr1(long t, long l) {
  if (l == maxlevel) {
    a[n] = t;
    n++;
  } else {
    pr1(t * 10 + 4, l + 1);
    pr1(t * 10 + 7, l + 1);
  }
}
int main() {
  a[0] = 0;
  cin >> p1 >> p2 >> v1 >> v2 >> k;
  for (maxlevel = 2; maxlevel <= 10; maxlevel++) pr1(0, 1);
  a[n] = 1000000001;
  n++;
  for (long i = 1; i < n - k; i++) {
    res(a[i - 1] + 1, a[i], a[i + k - 1], a[i + k] - 1);
  }
  cout.precision(20);
  cout << double(ans) / (double((double(p2 - p1 + 1)) * (v2 - v1 + 1)));
  getchar();
  getchar();
  return 0;
}