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<long long> lu;
int K;
long long pl, pr, vl, vr;
const long long inf = 1LL << 60;
void dfs(long long x, int l) {
  lu.push_back(x);
  if (l >= 11) return;
  dfs(x * 10 + 4, l + 1);
  dfs(x * 10 + 7, l + 1);
}
inline double intersect(long long l1, long long r1, long long l2,
                        long long r2) {
  long long ll = max(l1, l2), rr = min(r1, r2);
  return ll > rr ? 0 : rr - ll + 1;
}
int main() {
  lu.clear();
  dfs(4, 1), dfs(7, 1);
  sort(lu.begin(), lu.end());
  while (cin >> pl >> pr >> vl >> vr >> K) {
    double a = 0, b = (double)(pr - pl + 1) * (vr - vl + 1);
    for (int i = 0; i < lu.size(); i++) {
      int j = i + K - 1;
      if (j >= lu.size()) break;
      long long l1 = i == 0 ? -inf : lu[i - 1] + 1;
      long long r1 = lu[i];
      long long l2 = lu[j];
      long long r2 = j == lu.size() - 1 ? inf : lu[j + 1] - 1;
      a += intersect(pl, pr, l1, r1) * intersect(vl, vr, l2, r2);
      a += intersect(vl, vr, l1, r1) * intersect(pl, pr, l2, r2);
      if (K == 1 && intersect(pl, pr, r1, r1) && intersect(vl, vr, l2, l2)) a--;
    }
    printf("%.18lf\n", a / b);
  }
  return 0;
}