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;
long long pl, pr, vl, vr, K;
vector<long long> C;
double cut_p(long long L, long long R) {
  L = max(L, pl);
  R = min(R, pr);
  if (L > R) return 0;
  return (R - L + 1.0) / (pr - pl + 1.0);
}
double cut_v(long long L, long long R) {
  L = max(L, vl);
  R = min(R, vr);
  if (L > R) return 0;
  return (R - L + 1.0) / (vr - vl + 1.0);
}
int main() {
  cin >> pl >> pr >> vl >> vr >> K;
  for (int i = 1; i <= 9; ++i)
    for (int msk = 0; msk < (1 << i); ++msk) {
      long long T = 0;
      for (int j = 0; j < i; ++j) {
        T = 10 * T;
        if (msk & (1 << j))
          T += 7;
        else
          T += 4;
      }
      C.push_back(T);
    }
  C.push_back(-1);
  C.push_back(1000000002);
  sort(C.begin(), C.end());
  double ans = 0;
  for (int i = 1; i < C.size(); ++i) {
    int j = i + K - 1;
    if (j >= ((int)(C).size()) - 1) break;
    ans += cut_p(C[i - 1] + 1, C[i]) * cut_v(C[j], C[j + 1] - 1);
    ans += cut_v(C[i - 1] + 1, C[i]) * cut_p(C[j], C[j + 1] - 1);
  }
  if (K == 1) {
    double sub = 1.0 / (pr - pl + 1);
    sub /= (vr - vl + 1);
    for (int i = 0; i < C.size(); ++i)
      if (pl <= C[i] && C[i] <= pr && vl <= C[i] && C[i] <= vr) ans -= sub;
  }
  printf("%.12lf\n", ans);
  return 0;
}