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 int N = 100010;
long long next(long long x) {
  string s = to_string(x);
  int i = 0;
  while (i < s.size() && (s[i] == '4' || s[i] == '7')) i++;
  if (i >= s.size()) {
    i = s.size() - 1;
    while (i >= 0 && s[i] == '7') i--;
    if (i >= 0) {
      s[i] = '7';
      i++;
      while (i < s.size()) s[i++] = '4';
      return stoll(s);
    }
    string res = "";
    while (res.size() <= s.size()) res += "4";
    return stoll(res);
  }
  if (s[i] < '4') {
    while (i < s.size()) s[i++] = '4';
    return stoll(s);
  }
  if (s[i] < '7') {
    s[i] = '7';
    i++;
    while (i < s.size()) s[i++] = '4';
    return stoll(s);
  }
  i--;
  while (i >= 0 && s[i] == '7') i--;
  if (i >= 0) {
    s[i] = '7';
    i++;
    while (i < s.size()) s[i++] = '4';
    return stoll(s);
  }
  string res = "";
  while (res.size() <= s.size()) res += '4';
  return stoll(res);
}
bool overlap(long long a, long long b, long long c, long long d) {
  if (b < c || d < a) return false;
  return true;
}
int main() {
  long long pl, pr, vl, vr, k, x, y, px, py, res = 0, kk = 1;
  cin >> pl >> pr >> vl >> vr >> k;
  x = y = next(pl - 1);
  while (y < vr && kk < k) y = next(y), kk++;
  if (y <= vr && kk == k) {
    px = pl - 1;
    py = max(vl - 1, y - 1);
    while (y <= vr && px < pr) {
      if (min(next(y) - 1, vr) > py && min(x, pr) > px) {
        res += (min(x, pr) - px) * (min(next(y) - 1, vr) - py);
        px = min(x, pr);
        py = min(next(y) - 1, vr);
      } else
        px = max(x, pl - 1), py = max(y, vl - 1);
      y = next(y);
      x = next(x);
    }
  }
  swap(pl, vl);
  swap(pr, vr);
  kk = 1;
  x = y = next(pl - 1);
  while (y < vr && kk < k) y = next(y), kk++;
  if (y <= vr && kk == k) {
    px = pl - 1;
    py = max(vl - 1, y - 1);
    while (y <= vr && px < pr) {
      if (min(next(y) - 1, vr) > py && min(x, pr) > px) {
        res += (min(x, pr) - px) * (min(next(y) - 1, vr) - py);
        if (overlap(px + 1, min(x, pr), py + 1, min(next(y) - 1, vr))) res--;
        px = min(x, pr);
        py = min(next(y) - 1, vr);
      } else
        px = max(x, pl - 1), py = max(y, vl - 1);
      y = next(y);
      x = next(x);
    }
  }
  printf("%0.10lf",
         min(1.0 * res / (1.0 * (pr - pl + 1) * (vr - vl + 1)), 1.0));
  return 0;
}