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> a;
void pre() {
  a.push_back(0);
  for (int i = 0; i < 512; i++) {
    a.push_back(10 * a[i] + 4);
    a.push_back(10 * a[i] + 7);
  }
}
long long num(long long a, long long b, long long c, long long d) {
  return (min(b, d) >= max(a, c)) ? min(b, d) - max(a, c) + 1 : 0;
}
int main() {
  pre();
  long long pl, pr, vl, vr, p, v, k, sum = 0, first, second;
  scanf("%lld %lld %lld %lld %lld", &pl, &pr, &vl, &vr, &k);
  int low = 0;
  for (low = 0; low + k < 1023; low++) {
    first = num(a[low] + 1, a[low + 1], pl, pr);
    second = num(a[low + k], a[low + k + 1] - 1, vl, vr);
    sum += (first * second);
  }
  for (low = 0; low + k < 1023; low++) {
    first = num(a[low] + 1, a[low + 1], vl, vr);
    second = num(a[low + k], a[low + k + 1] - 1, pl, pr);
    sum += (first * second);
  }
  if (k == 1) {
    for (int i = 0; i < 1023; i++) {
      if (pl <= a[i] && pr >= a[i] && vl <= a[i] && vr >= a[i]) sum--;
    }
  }
  double total = (pr - pl + 1) * (vr - vl + 1);
  printf("%.15lf\n", sum / total);
  return 0;
}