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> Q;
void gen(long long n) {
  if (n > 4444444444LL) return;
  if (n) Q.push_back(n);
  gen(n * 10 + 7);
  gen(n * 10 + 4);
}
int main() {
  gen(0);
  sort(Q.begin(), Q.end());
  int pl, pr, vl, vr, k;
  scanf("%d%d%d%d%d", &pl, &pr, &vl, &vr, &k);
  double res = 0, inter = 0;
  ;
  for (int www = 0, n_ = (2); www < n_; www++) {
    for (int i = 0, n_ = ((int)(Q).size()); i < n_; i++) {
      long long lr = min((long long)pr, Q[i]);
      long long ll, rl, rr;
      if (i > 0) {
        ll = max((long long)pl, Q[i - 1] + 1);
      } else
        ll = pl;
      if (lr < ll) continue;
      double p1 = 0, p2 = 0;
      if (lr >= ll) {
        p1 = (lr - ll + 1);
      } else
        p1 = 0;
      int np = i + k - 1;
      if (np >= (int)(Q).size())
        p2 = 0;
      else {
        rl = max((long long)vl, Q[np]);
        rr;
        if (np + 1 < (int)(Q).size()) {
          rr = min(Q[np + 1] - 1, (long long)vr);
        } else
          rr = vr;
        if (rr >= rl)
          p2 = (rr - rl + 1);
        else
          p2 = 0;
      }
      if (rl <= rr && ll <= lr) {
        long long mm = max(ll, rl);
        long long mx = min(lr, rr);
        if (mm <= mx) inter += mx - mm + 1;
      }
      res += p1 * p2;
    }
    swap(vl, pl);
    swap(vr, pr);
  }
  res -= inter / 2;
  res = res / ((double)(vr - vl + 1) * (double)(pr - pl + 1));
  printf("%.15lf\n", res);
  return 0;
}