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 s1, e1, s2, e2, k;
vector<int> lucky;
inline void go(int now, int poi) {
  if (poi == 10) return;
  if (poi != 0) lucky.push_back(now);
  go(now * 10 + 4, poi + 1);
  go(now * 10 + 7, poi + 1);
}
inline double prob(int s1, int e1, int s2, int e2) {
  int S = max(s1, s2);
  int E = min(e1, e2);
  if (S > E) return 0.0;
  return ((double)E - S + 1.0) / ((double)e1 - s1 + 1.0);
}
int main() {
  cin >> s1 >> e1 >> s2 >> e2 >> k;
  go(0, 0);
  sort(lucky.begin(), lucky.end());
  int len = (int)lucky.size();
  double res = 0.0;
  for (int i = 0; i < len; i++) {
    int j = i + k - 1;
    if (j >= len) break;
    int beg1 = (i == 0) ? 1 : lucky[i - 1] + 1;
    int fin1 = lucky[i];
    int beg2 = lucky[j];
    int fin2 = (j == len - 1) ? 1000 * 1000 * 1000 : lucky[j + 1] - 1;
    if (k != 1) {
      res += prob(s1, e1, beg1, fin1) * prob(s2, e2, beg2, fin2);
      res += prob(s2, e2, beg1, fin1) * prob(s1, e1, beg2, fin2);
    } else {
      res += prob(s1, e1, beg1, fin1) * prob(s2, e2, beg2, fin2);
      res += prob(s2, e2, beg1, fin1) * prob(s1, e1, beg2, fin2);
      res -= prob(s1, e1, fin1, fin1) * prob(s2, e2, beg2, beg2);
    }
  }
  cout << fixed << setprecision(12) << res << endl;
  return 0;
}