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>
int pl, pr, vl, vr, k;
long long Calculate(int al, int ar, int bl, int br) {
  if (pl > al) al = pl;
  if (pr < ar) ar = pr;
  if (vl > bl) bl = vl;
  if (vr < br) br = vr;
  if (ar < al || br < bl) return 0;
  return (long long)(ar - al + 1) * (br - bl + 1);
}
int main() {
  int lucky[1024], n = 0;
  lucky[0] = 0;
  ++n;
  for (int i = 0; i < n; ++i) {
    if (lucky[i] > 100000000) continue;
    lucky[n++] = lucky[i] * 10 + 4;
    lucky[n++] = lucky[i] * 10 + 7;
  }
  lucky[n] = 1000000001;
  while (~scanf("%d %d %d %d %d", &pl, &pr, &vl, &vr, &k)) {
    long long total = (long long)(pr - pl + 1) * (vr - vl + 1), ways = 0;
    for (int i = 1; i <= n - k; ++i) {
      ways += Calculate(lucky[i - 1] + 1, lucky[i], lucky[i + k - 1],
                        lucky[i + k] - 1);
      ways += Calculate(lucky[i + k - 1], lucky[i + k] - 1, lucky[i - 1] + 1,
                        lucky[i]);
      if (k == 1) ways -= Calculate(lucky[i], lucky[i], lucky[i], lucky[i]);
    }
    printf("%0.10lf\n", (double)ways / total);
  }
  return 0;
}