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 cmp(const void *a, const void *b) { return *((int *)a) - *((int *)b); }
int maximum(int a, int b) {
  if (a > b) {
    return a;
  } else {
    return b;
  }
}
int minimum(int a, int b) {
  if (a < b) {
    return a;
  } else {
    return b;
  }
}
int main() {
  int a[2000];
  int p = 0, q = 1, x = 1, i, j;
  int pl, pr, vl, vr, k;
  long long c = 0, d;
  a[0] = 0;
  for (i = 0; i < 9; i++) {
    for (j = p; j < q; j++) {
      a[x++] = a[j] * 10 + 4;
      a[x++] = a[j] * 10 + 7;
    }
    p = q;
    q = x;
  }
  qsort(a, x, sizeof(int), cmp);
  a[x] = 1000000001;
  scanf("%d %d %d %d %d", &pl, &pr, &vl, &vr, &k);
  d = (long long)(pr - pl + 1) * (vr - vl + 1);
  for (i = 1; i <= x - k; i++) {
    if (pl <= a[i] && vr >= a[i + k - 1]) {
      int s1 = minimum(a[i], pr);
      int s2 = maximum(a[i - 1] + 1, pl);
      int t1 = maximum(a[i + k - 1], vl);
      int t2 = minimum(a[i + k] - 1, vr);
      if (s1 - s2 >= 0 && t2 - t1 >= 0) {
        c += (long long)(s1 - s2 + 1) * (t2 - t1 + 1);
      }
    }
    if (vl <= a[i] && pr >= a[i + k - 1]) {
      int s1 = minimum(a[i], vr);
      int s2 = maximum(a[i - 1] + 1, vl);
      int t1 = maximum(a[i + k - 1], pl);
      int t2 = minimum(a[i + k] - 1, pr);
      if (s1 - s2 >= 0 && t2 - t1 >= 0) {
        c += (long long)(s1 - s2 + 1) * (t2 - t1 + 1);
      }
    }
  }
  if (k == 1) {
    for (i = 1; i < x; i++) {
      if (pl <= a[i] && pr >= a[i] && vl <= a[i] && vr >= a[i]) c--;
    }
  }
  printf("%.20lf\n", (double)c / d);
  return 0;
}