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 long long P = (1000 * 1000 * 1000) + 7;
const int arrn = 4e5 + 3;
long long comp(long long l, long long r, long long pl, long long pr) {
  return max((long long)0, (min(r, pr) - max(l, pl) + 1));
}
void solve() {
  long long pl, pr, vl, vr, k;
  cin >> pl >> pr >> vl >> vr >> k;
  vector<long long> temp;
  for (int i = 1; i < 12; i++) {
    for (int j = 0; j < (1 << i); j++) {
      long long num = 0;
      long long pow = 1;
      for (int k = 0; k < i; k++) {
        if (j & (1 << k))
          num += 7 * pow;
        else
          num += 4 * pow;
        pow *= 10;
      }
      temp.push_back(num);
    }
  }
  long long out = 0;
  for (int i = 0; i + k - 1 < temp.size(); i++) {
    long long x1;
    long long x2;
    x1 = comp(((i == 0) ? 1 : (temp[i - 1] + 1)), temp[i], pl, pr);
    x2 = comp(temp[i + k - 1], ((i == temp.size()) ? P : (temp[i + k] - 1)), vl,
              vr);
    out += x1 * x2;
  }
  for (int i = 0; i + k - 1 < temp.size(); i++) {
    long long x1;
    long long x2;
    x1 = comp(((i == 0) ? 1 : (temp[i - 1] + 1)), temp[i], vl, vr);
    x2 = comp(temp[i + k - 1], ((i == temp.size()) ? P : (temp[i + k] - 1)), pl,
              pr);
    out += x1 * x2;
  }
  long long total = (pr - pl + 1) * (vr - vl + 1);
  if (k == 1) {
    long long x1 = max(pl, vl);
    long long x2 = min(pr, vr);
    if (x1 <= x2) {
      int temp1 = (upper_bound(temp.begin(), temp.end(), x2) -
                   lower_bound(temp.begin(), temp.end(), x1));
      out -= temp1;
    }
  }
  double ans = (double)out / total;
  cout << setprecision(10) << ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int T;
  T = 1;
  while (T--) {
    solve();
  }
  return 0;
}