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;
template <typename T>
using minpq = priority_queue<T, vector<T>, greater<T>>;
long long pl, pr, vl, vr;
int k;
long long inter(long long a, long long b, long long c, long long d) {
  return max(0LL, min(b, d) - max(a, c) + 1);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> pl >> pr >> vl >> vr >> k;
  long double ans = 0;
  vector<long long> lucky;
  for (int len = (0); len < (11); len++) {
    for (int x = (0); x < ((1 << len)); x++) {
      long long y = 0;
      for (int i = (0); i < (len); i++) {
        if ((x >> i) & 1) {
          y = y * 10 + 4;
        } else {
          y = y * 10 + 7;
        }
      }
      lucky.push_back(y);
    }
  }
  sort((lucky).begin(), (lucky).end());
  int n = ((int)(lucky).size());
  for (int i = (1); i < (n - k); i++) {
    int j = i + k;
    ans += 1.0 * inter(pl, pr, lucky[i - 1] + 1, lucky[i]) *
           inter(vl, vr, lucky[j - 1], lucky[j] - 1);
    ans += 1.0 * inter(vl, vr, lucky[i - 1] + 1, lucky[i]) *
           inter(pl, pr, lucky[j - 1], lucky[j] - 1);
    if (k == 1 && pl <= lucky[i] && lucky[i] <= pr && vl <= lucky[i] &&
        lucky[i] <= vr)
      ans--;
  }
  ans /= 1.0 * (pr - pl + 1) * (vr - vl + 1);
  cout << fixed << setprecision(11) << ans << '\n';
}