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> lucky;
void genall(long long x = 0) {
  if (x > 1e9) return;
  lucky.push_back(x);
  genall(x * 10 + 4);
  genall(x * 10 + 7);
}
void solve() {
  int k;
  int a, b, c, d;
  cin >> a >> b >> c >> d >> k;
  pair<int, int> prange = {a, b}, vrange = {c, d};
  genall();
  sort(lucky.begin(), lucky.end());
  lucky.push_back(1e9 + 2);
  auto numofways = [&](pair<int, int> range, pair<int, int> request) {
    request.first = max(request.first, range.first);
    request.second = min(request.second, range.second);
    long long ans = max(request.second - request.first + 1, 0);
    return ans;
  };
  pair<int, int> intersection = {max(prange.first, vrange.first),
                                 min(prange.second, vrange.second)};
  long long ways = 0;
  for (int i = 1; i < lucky.size() - k; i++) {
    int j = i + k - 1;
    pair<int, int> prev_range = {lucky[i - 1] + 1, lucky[i]};
    pair<int, int> next_range = {lucky[j], lucky[j + 1] - 1};
    long long added = 0;
    added += numofways(prange, prev_range) * numofways(vrange, next_range);
    added += numofways(vrange, prev_range) * numofways(prange, next_range);
    if (i == j and prange.first <= lucky[i] and lucky[i] <= prange.second and
        vrange.first <= lucky[i] and lucky[i] <= vrange.second and added) {
      added--;
    }
    ways += added;
  }
  long long totalways = 1ll * (prange.second - prange.first + 1) *
                        (vrange.second - vrange.first + 1);
  long double prob = (long double)ways / totalways;
  cout << fixed << setprecision(10) << prob << "\n";
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  solve();
  return 0;
}