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;
long long inter(long long ax, long long ay, long long bx, long long by) {
  return max(0ll, min(ay, by) - max(ax, bx) + 1);
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL);
  ;
  long long pl, pr, vl, vr, k;
  cin >> pl >> pr >> vl >> vr >> k;
  vector<long long> a;
  a.push_back(0);
  for (int n = 1; n <= 10; n++) {
    for (int i = 0; i < (1 << n); i++) {
      string s = "";
      for (int j = 0; j < n; j++) {
        s += (((i >> j) & 1) ? '4' : '7');
      }
      a.push_back(stoll(s));
    }
  }
  a.push_back(100000000000);
  int n = a.size();
  sort(a.begin(), a.end());
  long long ans = 0;
  for (int i = 1; i + k - 1 < n - 1; i++) {
    ans += inter(pl, pr, a[i - 1] + 1, a[i]) *
               inter(vl, vr, a[i + k - 1], a[i + k] - 1) +
           inter(vl, vr, a[i - 1] + 1, a[i]) *
               inter(pl, pr, a[i + k - 1], a[i + k] - 1);
    if (k == 1 && a[i] >= pl && a[i] <= pr && a[i] >= vl && a[i] <= vr) ans--;
  }
  printf("%.9f\n", (double)ans / ((pr - pl + 1) * (vr - vl + 1)));
  return 0;
}