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;
const double eps = 1e-9;
long long num[2000];
int cnum;
int main() {
  cnum = 0;
  num[cnum++] = 0;
  for (int i = 0; i < cnum; i++) {
    num[cnum++] = num[i] * 10 + 4;
    num[cnum++] = num[i] * 10 + 7;
    if (num[cnum - 1] > 1000000000) break;
  }
  long long a, b, c, d;
  int k;
  cin >> a >> b >> c >> d >> k;
  long long ans = 0;
  int tmp = lower_bound(num, num + cnum, a) - num;
  int tp = upper_bound(num, num + cnum, d) - num - 1;
  int ok = 0;
  if (tp - tmp + 1 >= k) {
    ok = 1;
    for (int i = tmp; i + k - 1 <= tp; ++i) {
      double tt = min(num[i], b) - max(a, num[i - 1] + 1) + 1;
      double tq = i + k <= cnum
                      ? min(d, num[i + k] - 1) - max(c, num[i + k - 1]) + 1
                      : d - max(c, num[i + k - 1]) + 1;
      if (tt * tq > 0 && tt > 0 && tq > 0) ans += tt * tq;
    }
  }
  tmp = lower_bound(num, num + cnum, c) - num;
  tp = upper_bound(num, num + cnum, b) - num - 1;
  if (tp - tmp + 1 >= k) {
    for (int i = tmp; i + k - 1 <= tp; ++i) {
      if (ok && k == 1) ans--;
      double tt = min(num[i], d) - max(c, num[i - 1] + 1) + 1;
      double tq = i + k <= cnum
                      ? min(b, num[i + k] - 1) - max(a, num[i + k - 1]) + 1
                      : b - max(a, num[i + k - 1]) + 1;
      if (tt * tq > 0 && tt > 0 && tq > 0) ans += tt * tq;
    }
  }
  cout << setiosflags(ios::fixed) << setprecision(12)
       << (double)ans / (b - a + 1) / (d - c + 1) << endl;
  return 0;
}