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> getLucky() {
  vector<int> lucky;
  lucky.push_back(0);
  for (int len = (1); len <= (9); ++len)
    for (int mask = (0); mask <= ((1 << len) - 1); ++mask) {
      long long int x = 0, pow = 1;
      for (int i = (0); i <= (len - 1); ++i) {
        if (mask & (1 << i))
          x += pow * 7;
        else
          x += pow * 4;
        pow *= 10;
      }
      lucky.push_back(x);
    }
  lucky.push_back(1000000001);
  return lucky;
}
double len(double l, double r) {
  if (r >= l)
    return (r - l + 1);
  else
    return 0;
}
double solve(int p[], int X[], bool left_open) {
  int i[2] = {max(p[0], X[0]), min(p[1], X[1])};
  if (i[0] == X[0] && left_open)
    i[0]++;
  else if (i[1] == X[1] && !left_open)
    i[1]--;
  double ans = len(i[0], i[1]);
  ans /= len(p[0], p[1]);
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  int p[2], v[2], k;
  for (int i = (0); i <= (1); ++i) cin >> p[i];
  for (int i = (0); i <= (1); ++i) cin >> v[i];
  cin >> k;
  vector<int> lucky = getLucky();
  double ans = 0;
  for (int i = (1); i <= ((int)(lucky).size() - k - 1); ++i) {
    int L[2] = {lucky[i - 1], lucky[i]},
        R[2] = {lucky[i + k - 1], lucky[i + k]};
    double prob =
        solve(p, L, 1) * solve(v, R, 0) + solve(p, R, 0) * solve(v, L, 1);
    int in[2] = {max(p[0], v[0]), min(p[1], v[1])};
    if (k == 1 && lucky[i] >= in[0] && lucky[i] <= in[1])
      prob -= 1.0 / (len(p[0], p[1]) * len(v[0], v[1]));
    ans += prob;
  }
  cout.precision(12);
  cout << fixed << ans;
  return 0;
}