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> ls;
void rec(long long cur) {
  if (cur > 1e9) return;
  ls.push_back(cur);
  rec(cur * 10 + 4);
  rec(cur * 10 + 7);
}
long long isII(long long l, long long r, long long L, long long R) {
  return max(min(r, R) - max(l, L) + 1, 0LL);
}
int main() {
  rec(0);
  ls.push_back(INT_MAX);
  sort(ls.begin(), ls.end());
  long long l, r, L, R, k;
  cin >> l >> r >> L >> R >> k;
  long long cnt = 0;
  for (int i = 1; i < ls.size() - 1; ++i) {
    int j = i + k - 1;
    if (j < ls.size() - 1) {
      {
        long long A = isII(ls[i - 1] + 1, ls[i], l, r);
        long long B = isII(ls[j], ls[j + 1] - 1, L, R);
        cnt += A * B;
      }
      {
        long long A = isII(ls[i - 1] + 1, ls[i], L, R);
        long long B = isII(ls[j], ls[j + 1] - 1, l, r);
        cnt += A * B;
      }
      if (k == 1) {
        if (l <= ls[i] && ls[i] <= r && L <= ls[i] && ls[i] <= R) --cnt;
      }
    }
  }
  double ans = 1.0 * cnt / ((r - l + 1) * (R - L + 1));
  printf("%0.10lf\n", ans);
  return 0;
}