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;
double get(pair<int, int> a, pair<int, int> b) {
  double ret = 0;
  int x = max(a.first, b.first);
  int y = min(a.second, b.second);
  if (x > y)
    return 0;
  else
    ret = (y - x + 1);
  return ret;
}
int main() {
  int i, n, j, k, till, now, pl, pr, vl, vr, Ka, siz;
  scanf("%d %d %d %d %d", &pl, &pr, &vl, &vr, &Ka);
  vector<int> v;
  pair<int, int> p1, p2;
  p1 = make_pair(pl, pr);
  p2 = make_pair(vl, vr);
  for (k = 1; k <= 9; k++) {
    till = (1 << k) - 1;
    for (i = 0; i <= till; i++) {
      now = 0;
      for (j = 0; j < k; j++) {
        now *= 10;
        if ((1 << j) & i)
          now += 7;
        else
          now += 4;
      }
      v.push_back(now);
    }
  }
  siz = v.size();
  sort(v.begin(), v.end());
  double lob, hor;
  hor = fabs(pr - pl + 1);
  hor *= fabs((double)(vr - vl + 1));
  lob = 0;
  for (i = 0; i <= (siz - Ka); i++) {
    j = i + Ka - 1;
    pair<int, int> l, r;
    if (i == 0)
      l = make_pair(1, v[i]);
    else
      l = make_pair(v[i - 1] + 1, v[i]);
    if (j == (siz - 1))
      r = make_pair(v[j], 1000000000);
    else
      r = make_pair(v[j], v[j + 1] - 1);
    if (l.second == r.first) {
      int f = l.second;
      r.first++;
      lob += get(l, p1) * get(r, p2);
      lob += get(l, p2) * get(r, p1);
      l.second--;
      r.first = f;
      r.second = f;
      lob += get(l, p1) * get(r, p2);
      lob += get(l, p2) * get(r, p1);
      if (vl <= f && f <= vr && pl <= f && f <= pr) lob++;
    } else {
      lob += get(l, p1) * get(r, p2);
      lob += get(l, p2) * get(r, p1);
    }
  }
  printf("%.12lf\n", lob / hor);
  return 0;
}