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<int> posibles(int a, int b) {
  int i, j, n, m, c, v[2] = {4, 7};
  vector<int> res;
  for (n = (1); n < (10); n++) {
    m = 1 << n;
    for (i = (0); i < (m); i++) {
      c = 0;
      for (j = (0); j < (n); j++) {
        c = c * 10 + v[(i & (1 << j)) >> j];
      }
      if (a <= c && c <= b) {
        res.push_back(c);
      }
    }
  }
  sort(res.begin(), res.end());
  return res;
}
int read() {
  int n;
  char c;
  bool neg;
  n = 0;
  neg = false;
  for (c = getchar(); !(c >= '0' && c <= '9') && c != '-'; c = getchar())
    ;
  if (c == '-') {
    neg = true;
    c = getchar();
  }
  while ((c >= '0' && c <= '9')) {
    n = n * 10 + c - '0';
    c = getchar();
  }
  return neg ? -n : n;
}
int interseccion(int x1, int y1, int x2, int y2) {
  pair<int, int> res;
  if ((x2 <= x1 && x1 <= y2) || (x2 <= y1 && y1 <= y2) ||
      (x1 <= x2 && x2 <= y1) || (x1 <= y2 && y2 <= y1)) {
    res = make_pair(min(y1, y2), max(x1, x2));
    if (res.first > res.second) {
      swap(res.first, res.second);
    }
    return res.second - res.first + 1;
  }
  return 0;
}
int main() {
  int pl, pr, vl, vr, k, i, j, t1, t2, t3, t4, x1, y1, x2, y2;
  long long int temp;
  double nm;
  vector<int> v;
  cin >> pl >> pr >> vl >> vr >> k;
  if (pl > vl) {
    swap(vl, pl);
    swap(vr, pr);
  }
  nm = double((long long int)(pr - pl + 1) * (long long int)(vr - vl + 1));
  v = posibles(min(pl, vl), max(pr, vr));
  temp = 0LL;
  if ((int)v.size() >= k) {
    for (i = 0, j = k - 1; j < (int)v.size(); i++, j++) {
      if (i == 0) {
        x1 = 1;
      } else {
        x1 = v[i - 1] + 1;
      }
      y1 = v[i];
      x2 = v[j];
      if (j + 1 == (int)v.size()) {
        y2 = 1000000000;
      } else {
        y2 = v[j + 1] - 1;
      }
      t1 = interseccion(x1, y1, vl, vr);
      t2 = interseccion(x2, y2, pl, pr);
      t3 = interseccion(x1, y1, pl, pr);
      t4 = interseccion(x2, y2, vl, vr);
      temp += ((long long int)t1 * (long long int)t2) +
              ((long long int)t3 * (long long int)t4);
      if (k == 1 && pl <= v[i] && v[i] <= pr && vl <= v[i] && v[i] <= vr) {
        temp--;
      }
    }
  }
  printf("%.12lf\n", double(temp) / nm);
  return 0;
}