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;
long long n, m, i, j, x, y, g, h, k, d, ii, res, q, w, e;
long long a[1000], b[2000][20];
long long func(int j) {
  long long res = 0, i;
  for (i = 0; b[j][i]; ++i) res = res * 10 + b[j][i];
  return res;
}
int main() {
  a[0] = 4;
  b[0][0] = 4;
  n = m = 1;
  while (n < 10) {
    j = n - 1;
    while (j >= 0 && a[j] == 7) j--;
    if (j < 0) {
      n++;
      for (j = 0; j < n; ++j) a[j] = 4;
    } else {
      a[j] = 7;
      for (j++; j < n; ++j) a[j] = 4;
    }
    for (j = 0; j < n; ++j) b[m][j] = a[j];
    m++;
  }
  cin >> x >> y >> g >> h >> k;
  m--;
  for (ii = 0; ii < m; ++ii) {
    d = func(ii);
    if (d >= x) break;
  }
  i = x;
  while (i <= y) {
    q = func(ii);
    if (ii + k - 1 < m) {
      w = func(ii + k - 1);
      e = func(ii + k);
      if ((g <= w && w <= h) || (w <= g && e >= g)) {
        res = res + ((((q) < (y)) ? (q) : (y)) - i + 1) *
                        ((((h) < (e - 1)) ? (h) : (e - 1)) -
                         (((w) > (g)) ? (w) : (g)) + 1);
      }
    }
    i = q + 1;
    ii++;
  }
  swap(g, x);
  swap(y, h);
  for (ii = 0; ii < m; ++ii) {
    d = func(ii);
    if (d >= x) break;
  }
  i = x;
  while (i <= y) {
    q = func(ii);
    if (ii + k - 1 < m) {
      w = func(ii + k - 1);
      e = func(ii + k);
      if ((g <= w && w <= h) || (w <= g && e >= g)) {
        res = res + ((((q) < (y)) ? (q) : (y)) - i + 1) *
                        ((((h) < (e - 1)) ? (h) : (e - 1)) -
                         (((w) > (g)) ? (w) : (g)) + 1);
        if ((((q) < (y)) ? (q) : (y)) == (((w) > (g)) ? (w) : (g))) res--;
      }
    }
    i = q + 1;
    ii++;
  }
  if (res > ((y - x + 1) * (h - g + 1)))
    cout << 1;
  else
    printf("%.9lf", (double)(res) / ((y - x + 1) * (h - g + 1)));
  return 0;
}