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> luck;
int main() {
  long long p1, p2, v1, v2, k;
  cin >> p1 >> p2 >> v1 >> v2 >> k;
  for (int k = 0; k < 11; k++)
    for (int i = 0; i < (1 << k); i++) {
      long long p = 0;
      for (int j = 0; j < k; j++)
        if (i & (1 << j))
          p = p * 10 + 7;
        else
          p = p * 10 + 4;
      luck.push_back(p);
    }
  sort(luck.begin(), luck.end());
  long double ans = 0;
  for (int i = 1; i < luck.size() - k; i++) {
    long long a = luck[i], b = luck[i + k - 1], c = luck[i - 1] + 1,
              d = luck[i + k] - 1;
    if (k != 1) {
      if (min(a, p2) + 1 >= max(p1, c) && min(d, v2) + 1 >= max(b, v1))
        ans += 1.0 * (min(a, p2) - max(p1, c) + 1) / (p2 - p1 + 1) * 1.0 *
               (min(d, v2) - max(b, v1) + 1) / (v2 - v1 + 1);
      if (min(a, v2) + 1 >= max(v1, c) && min(d, p2) + 1 >= max(b, p1))
        ans += 1.0 * (min(a, v2) - max(v1, c) + 1) / (v2 - v1 + 1) * 1.0 *
               (min(d, p2) - max(b, p1) + 1) / (p2 - p1 + 1);
    } else {
      if (min(a, p2) + 1 >= max(p1, c) && min(d, v2) + 1 >= max(b, v1))
        ans += 1.0 * (min(a, p2) - max(p1, c) + 1) / (p2 - p1 + 1) * 1.0 *
               (min(d, v2) - max(b, v1) + 1) / (v2 - v1 + 1);
      if (min(a, v2) + 1 >= max(v1, c) && min(d, p2) + 1 >= max(b, p1))
        ans += 1.0 * (min(a, v2) - max(v1, c) + 1) / (v2 - v1 + 1) * 1.0 *
               (min(d, p2) - max(b, p1) + 1) / (p2 - p1 + 1);
      if (min(a, v2) >= max(v1, c) && min(d, p2) >= max(b, p1) &&
          min(a, p2) >= max(p1, c) && min(d, v2) >= max(b, v1))
        ans -= 1.0 / (p2 - p1 + 1) * 1.0 / (v2 - v1 + 1);
    }
  }
  cout << fixed << setprecision(12) << ans << endl;
}