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>
int a, b, c, d, k;
int get(int a, int b, int l, int r) {
  if (a > r || b < l) {
    return 0;
  }
  return std::min(r, b) - std::max(l, a) + 1;
}
int main() {
  scanf("%d%d%d%d%d", &a, &b, &c, &d, &k);
  std::queue<int> q;
  std::vector<int> vt(0);
  q.push(0);
  while (q.size()) {
    int cur = q.front();
    q.pop();
    vt.push_back(cur);
    if (cur < (int)1e8) {
      q.push(cur * 10 + 4);
      q.push(cur * 10 + 7);
    }
  }
  vt.push_back(int(2e9));
  int size = vt.size() - 2;
  long long count = 0;
  for (int i = 1; i + k - 1 <= size; i++) {
    for (int j = 0; j <= 1; j++) {
      count += 1ll * get(a, b, vt[i - 1] + 1, vt[i]) *
               get(c, d, vt[i + k - 1], vt[i + k] - 1);
      if (a == c && b == d && a == b) {
      }
      std::swap(a, c);
      std::swap(b, d);
    }
    if (k == 1) {
      if (a <= vt[i] && b >= vt[i] && c <= vt[i] && d >= vt[i]) {
        count--;
      }
    }
  }
  printf("%.10f\n", 1.0 * count / ((b - a + 1.0) * (d - c + 1.0)));
  return 0;
}