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;
vector<long long> lucky;
int size;
void dfs(int i, long long num) {
  if (num > 1000000000) return;
  lucky.push_back(num);
  dfs(i + 1, num * 10 + 4);
  dfs(i + 1, num * 10 + 7);
}
long long func(long long x1, long long x2, long long y1, long long y2, int k) {
  int i, j;
  long long ans = 0;
  for ((i) = 0; (i) < (int)(size - k + 1); (i)++) {
    long long xlow = x1, xhigh = x2, ylow = y1, yhigh = y2;
    j = i + k - 1;
    if (i != 0) xlow = max(x1, lucky[i - 1] + 1);
    xhigh = min(x2, lucky[i]);
    ylow = max(y1, lucky[j]);
    if (j != (size - 1)) yhigh = min(lucky[j + 1] - 1, y2);
    if (xhigh >= xlow && yhigh >= ylow)
      ans += (xhigh - xlow + 1) * (yhigh - ylow + 1);
  }
  return ans;
}
int main() {
  int i, n, k;
  long long x1, x2, y1, y2;
  long long ans, total;
  dfs(0, 4);
  dfs(0, 7);
  sort(lucky.begin(), lucky.end());
  size = lucky.size();
  cin >> x1 >> x2 >> y1 >> y2 >> k;
  total = (x2 - x1 + 1) * (y2 - y1 + 1);
  ans = func(x1, x2, y1, y2, k) + func(y1, y2, x1, x2, k);
  if (k == 1)
    for ((i) = 0; (i) < (int)(size); (i)++)
      if (x1 <= lucky[i] && x2 >= lucky[i] && y1 <= lucky[i] && y2 >= lucky[i])
        ans--;
  cout << fixed << setprecision(12) << (double)ans / (double)total << endl;
  return 0;
}