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 main() {
  unsigned long long p1, p2, v1, v2, k, q[10000], first, last, x, res, i, x1,
      x2, y1, y2, slc1, slc2, sl1, sl2, xc1, xc2, yc1, yc2;
  cin >> p1 >> p2 >> v1 >> v2 >> k;
  q[1] = 4;
  q[2] = 7;
  first = 0;
  last = 2;
  while (first < last) {
    first++;
    x = q[first];
    if (x * 10 + 4 <= 1000000000) {
      last++;
      q[last] = x * 10 + 4;
    }
    if (x * 10 + 7 <= 1000000000) {
      last++;
      q[last] = x * 10 + 7;
    }
  }
  q[0] = 0;
  q[last + 1] = 1000000001;
  res = 0;
  k -= 1;
  for (i = 1; i <= last; i++) {
    if (i + k <= last) {
      x1 = q[i - 1] + 1;
      x2 = q[i];
      y2 = q[i + k + 1] - 1;
      y1 = q[i + k];
      xc1 = max(x1, p1);
      xc2 = min(x2, p2);
      yc1 = max(y1, v1);
      yc2 = min(y2, v2);
      if (xc2 >= xc1)
        sl1 = xc2 - xc1 + 1;
      else
        sl1 = 0;
      if (yc1 <= yc2)
        sl2 = yc2 - yc1 + 1;
      else
        sl2 = 0;
      res += sl1 * sl2;
      xc1 = max(x1, v1);
      xc2 = min(x2, v2);
      yc1 = max(y1, p1);
      yc2 = min(y2, p2);
      if (xc2 >= xc1)
        sl1 = xc2 - xc1 + 1;
      else
        sl1 = 0;
      if (yc1 <= yc2)
        sl2 = yc2 - yc1 + 1;
      else
        sl2 = 0;
      res += sl1 * sl2;
      if (k == 0 && q[i] >= p1 && q[i] <= p2 && q[i] >= v1 && q[i] <= v2) res--;
    }
  }
  sl1 = (p2 - p1 + 1);
  sl2 = (v2 - v1 + 1);
  slc1 = sl1 * sl2;
  printf("%.9f", 1.0 * res / slc1);
}