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 k, n1 = 0, n = 0, x;
long long int y, a[200005], b[200005], pl, pr, vl, vr, L, R;
long long int AC = 0;
long long int P_LEFT(int x) {
  if (!x)
    return (min(b[x], pr) - pl + 1);
  else
    return (min(b[x], pr) - max(pl, b[x - 1] + 1) + 1);
}
long long int V_LEFT(int x) {
  if (!x)
    return (min(b[x], vr) - vl + 1);
  else
    return (min(b[x], vr) - max(vl, b[x - 1] + 1) + 1);
}
long long int P_RIGHT(int x) {
  if (x == n - 1)
    return (pr - max(b[x], pl) + 1);
  else
    return (min(pr, b[x + 1] - 1) - max(b[x], pl) + 1);
}
long long int V_RIGHT(int x) {
  if (x == n - 1)
    return (vr - max(b[x], vl) + 1);
  else
    return (min(vr, b[x + 1] - 1) - max(b[x], vl) + 1);
}
int main() {
  scanf("%I64d%I64d%I64d%I64d%d", &pl, &pr, &vl, &vr, &k);
  L = pr - pl + 1;
  R = vr - vl + 1;
  for (int i = 1; i <= 9; i++)
    for (int j = 0; j < (1 << i); j++) {
      x = j;
      y = 0;
      for (int k = 0; k < i; k++) {
        if (x % 2 == 0)
          y = y * 10 + 4;
        else
          y = y * 10 + 7;
        x /= 2;
      }
      a[n1++] = y;
    }
  sort(a, a + n1);
  for (int i = 0; i < n1; i++)
    if (!i || a[i] != a[i - 1]) b[n++] = a[i];
  for (int i = 0; i + k - 1 < n; i++) {
    AC += max(0LL, P_LEFT(i)) * max(0LL, V_RIGHT(i + k - 1));
    AC += max(0LL, V_LEFT(i)) * max(0LL, P_RIGHT(i + k - 1));
    if (k == 1 && pl <= b[i] && b[i] <= pr && vl <= b[i] && b[i] <= vr) AC--;
  }
  printf("%.15lf\n", AC * 1.0 / (L * 1.0 * R));
  return 0;
}