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>
const long long INF = 1e18;
const int N = 1e5 + 5, inf = 1e9, mod = 1e9 + 7;
using namespace std;
long double pl, pr, vl, vr, k;
vector<long double> l;
void gen(long double x) {
  if (x * 10 + 4 <= max(pr, vr)) {
    if (x * 10 + 4 >= min(pl, vl)) l.push_back(x * 10 + 4);
    gen(x * 10 + 4);
  }
  if (x * 10 + 7 <= max(pr, vr)) {
    if (x * 10 + 7 >= min(pl, vl)) l.push_back(x * 10 + 7);
    gen(x * 10 + 7);
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> pl >> pr >> vl >> vr >> k;
  l.push_back(0);
  l.push_back(inf);
  gen(0);
  sort(l.begin(), l.end());
  if (l.size() - 2 < k) return cout << "0\n", 0;
  long double ans = 0.0;
  long double p = (pr - pl + 1), v = (vr - vl + 1);
  for (int i = 1; i < l.size() - k; i++) {
    if (pl <= l[i] && vr >= l[i + k - 1]) {
      long double tmp1 =
          max((long double)0,
              ((long double)min(pr, l[i]) - max(l[i - 1] + 1, pl) + 1) / p);
      long double tmp2 = max(
          (long double)0,
          ((long double)min(l[i + k] - 1, vr) - max(vl, l[i + k - 1]) + 1) / v);
      ans += tmp1 * tmp2;
    }
    if (vl <= l[i] && pr >= l[i + k - 1]) {
      long double tmp1 =
          max((long double)0,
              ((long double)min(vr, l[i]) - max(l[i - 1] + 1, vl) + 1) / v);
      long double tmp2 = max(
          (long double)0,
          ((long double)min(l[i + k] - 1, pr) - max(pl, l[i + k - 1]) + 1) / p);
      ans += tmp1 * tmp2;
    }
    if (pl <= l[i] && l[i] <= pr && vl <= l[i + k - 1] && l[i + k - 1] <= vr &&
        k == 1)
      ans -= (1.0 / p) * (1.0 / v);
  }
  cout << fixed << setprecision(10) << ans << '\n';
  return 0;
}