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;
int pl, pr, vl, vr, k;
vector<int> lucky_numbers;
void generate(long long x) {
  if (x >= 1e9) return;
  lucky_numbers.push_back(x);
  generate(x * 10 + 4);
  generate(x * 10 + 7);
}
long double find(int a, int b, int c, int d) {
  if (c > b || d < a) return 0;
  c = max(a, c), d = min(b, d);
  long double ONE = 1;
  return ONE * (d - c + 1) / (b - a + 1);
}
int main() {
  cin >> pl >> pr >> vl >> vr >> k;
  generate(4), generate(7);
  sort(lucky_numbers.begin(), lucky_numbers.end());
  long double ans = 0;
  for (int i = 0; i < lucky_numbers.size(); i++) {
    if (i + k - 1 > lucky_numbers.size()) break;
    int prv = i ? lucky_numbers[i - 1] + 1 : 1;
    int nxt =
        i + k - 1 == lucky_numbers.size() - 1 ? 1e9 : lucky_numbers[i + k] - 1;
    ans += find(pl, pr, prv, lucky_numbers[i]) *
           find(vl, vr, lucky_numbers[i + k - 1], nxt);
    ans += find(vl, vr, prv, lucky_numbers[i]) *
           find(pl, pr, lucky_numbers[i + k - 1], nxt);
    if (k == 1)
      ans -= find(vl, vr, lucky_numbers[i], lucky_numbers[i]) *
             find(pl, pr, lucky_numbers[i], lucky_numbers[i]);
  }
  cout << setprecision(10) << fixed << ans;
  return 0;
}