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 ar[2000], ind;
int init() {
  ar[0] = 0;
  ind = 1;
  char tmp[10];
  int ind2;
  for (int i = 1; i <= 9; i++) {
    for (int j = i; j >= 0; j--) {
      ind2 = 0;
      tmp[i] = 0;
      while (ind2 < j) tmp[ind2++] = '4';
      while (ind2 < i) tmp[ind2++] = '7';
      do {
        sscanf(tmp, "%d", &ar[ind++]);
      } while (next_permutation(tmp, tmp + i));
    }
  }
  ar[ind] = 1000000001;
  sort(ar, ar + ind);
  return 0;
}
int solve() {
  int pl, pr, vl, vr, k;
  scanf("%d %d %d %d %d", &pl, &pr, &vl, &vr, &k);
  int cur = 0;
  long long cnt = 0;
  int l1, l2, u1, u2, kak1, kak2;
  long long pp, vv;
  for (int i = 1; i < ind; i++) {
    if (i + k > ind) break;
    l1 = ar[i - 1] + 1;
    l2 = ar[i];
    u1 = ar[i + k - 1];
    u2 = ar[i + k] - 1;
    pp = min(l2, pr) - max(l1, pl) + 1;
    if (pp < 0) pp = 0;
    vv = min(u2, vr) - max(u1, vl) + 1;
    if (vv < 0) vv = 0;
    cnt += pp * vv;
    pp = min(u2, pr) - max(u1, pl) + 1;
    if (pp < 0) pp = 0;
    vv = min(l2, vr) - max(l1, vl) + 1;
    if (vv < 0) vv = 0;
    cnt += pp * vv;
    kak1 = max(l1, u1);
    kak2 = min(u2, l2);
    pp = min(kak2, pr) - max(kak1, pl) + 1;
    if (pp < 0) pp = 0;
    vv = min(kak2, vr) - max(kak1, vl) + 1;
    if (vv < 0) vv = 0;
    cnt -= pp * vv;
  }
  long long all1 = (pr - pl + 1);
  long long all2 = (vr - vl + 1);
  long long all = all1 * all2;
  double sol = 1.0 * cnt / all;
  if (sol > 1.0) sol = 1.0;
  if (sol < 0) sol = 0.0;
  printf("%.12lf\n", sol);
  return 0;
}
int main() {
  init();
  solve();
  return 0;
}