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 a[1022];
int minusa = 0;
long long calcul(int pl, int pr, int vl, int vr, int k, int i) {
  int l1 = max((i == 0 ? min(pl, a[i]) : a[i - 1] + 1), pl);
  int r1 = min(a[i], pr);
  int l2 = max(a[i + k - 1], vl);
  int r2 = min((i + k == 1022 ? vr : a[i + k] - 1), vr);
  if (r1 >= l1 && r2 >= l2) {
    if (r1 == l2) {
      minusa++;
    }
    return (long long)(r1 - l1 + 1) * (r2 - l2 + 1);
  }
  return 0;
}
int main() {
  queue<int> q;
  q.push(4);
  q.push(7);
  int d = 1;
  int cnt = 0;
  while (!q.empty() && d < 10) {
    int size = q.size();
    while (size--) {
      int plucky = q.front();
      a[cnt] = plucky;
      q.pop();
      if (d < 9) {
        q.push(plucky * 10 + 4);
        q.push(plucky * 10 + 7);
      }
      cnt++;
    }
    d++;
  }
  int pl, pr, vl, vr, k;
  scanf("%d %d %d %d %d", &pl, &pr, &vl, &vr, &k);
  long long trueSpace = 0;
  for (int i = 0; i <= 1022 - k; i++) {
    minusa = 0;
    trueSpace += calcul(pl, pr, vl, vr, k, i);
    trueSpace += calcul(vl, vr, pl, pr, k, i);
    if (minusa == 2) {
      trueSpace--;
    }
  }
  double res = (double)trueSpace / ((long long)(pr - pl + 1) * (vr - vl + 1));
  printf("%.13lf", res);
  return 0;
}