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;
int l[2000], ln;
void go(int a) {
  l[ln++] = a;
  if (a > 1e8) return;
  go(a * 10 + 4);
  go(a * 10 + 7);
}
void predo() {
  ln = 0;
  go(0);
  l[ln++] = 1000000000;
  sort(l, l + ln);
}
long long int len(int a, int b, int c, int d) {
  int k = min(b, d) - max(a, c) + 1;
  return k >= 0 ? k : 0;
}
int main() {
  predo();
  int l1, r1, l2, r2, k;
  scanf("%d%d%d%d%d", &l1, &r1, &l2, &r2, &k);
  long long int ans = 0;
  for (int i = 1; i + k < ln; i++) {
    ans += len(l[i - 1] + 1, l[i], l1, r1) *
           len(l[i + k - 1], l[i + k] - 1, l2, r2);
    ans += len(l[i - 1] + 1, l[i], l2, r2) *
           len(l[i + k - 1], l[i + k] - 1, l1, r1);
    if (k == 1 && len(l[i], l[i], l1, r1) == 1 && len(l[i], l[i], l2, r2) == 1)
      ans--;
  }
  printf("%.11lf\n", (double)ans / (r1 - l1 + 1) / (r2 - l2 + 1));
  return 0;
}