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 = 100005;
long long a[N];
int size;
void pre() {
  a[0] = 0;
  a[1] = 4;
  a[2] = 7;
  int s = 1, e = 2;
  size = 2;
  for (int i = 2; i <= 10; i++) {
    for (int j = s; j <= e; j++) {
      a[++size] = a[j] * 10 + 4;
      a[++size] = a[j] * 10 + 7;
    }
    s = e + 1;
    e = size;
  }
}
int main() {
  pre();
  long long pr, pl, vr, vl, k;
  scanf("%I64d %I64d %I64d %I64d %I64d", &pl, &pr, &vl, &vr, &k);
  double ans = 0;
  int s = 1;
  while (a[s] < pl) s++;
  while (a[s + k - 1] <= vr) {
    if (a[s + k] > vl && a[s - 1] < pr)
      ans += (min(a[s], pr) - max(pl, a[s - 1] + 1) + 1) *
             (min(a[s + k] - 1, vr) - max(vl, a[s + k - 1]) + 1);
    s++;
  }
  swap(pl, vl);
  swap(pr, vr);
  s = 1;
  while (a[s] < pl) s++;
  while (a[s + k - 1] <= vr) {
    if (a[s + k] > vl && a[s - 1] < pr)
      ans += (min(a[s], pr) - max(pl, a[s - 1] + 1) + 1) *
             (min(a[s + k] - 1, vr) - max(vl, a[s + k - 1]) + 1);
    s++;
  }
  if (k == 1) {
    for (int i = 1; i <= size; i++) {
      if (a[i] >= pl && a[i] <= pr && a[i] >= vl && a[i] <= vr) ans--;
    }
  }
  ans = ans / (pr - pl + 1) / (vr - vl + 1);
  printf("%.12f\n", ans);
  return 0;
}