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;
const double EPS = 1e-9;
const double PI = acos(-1.);
template <class T>
inline T sq(T x) {
  return x * x;
}
vector<int> luckys;
int _gen(int m, int n) {
  return n ? (m & 1 ? 7 : 4) + _gen(m >> 1, n - 1) * 10 : 0;
}
void gen() {
  for (int i = (1); i < (10); ++i) {
    for (int j = (0); j < (1 << i); ++j) {
      luckys.push_back(_gen(j, i));
    }
  }
}
double prob(int l1, int r1, int l2, int r2) {
  int l3 = max(l1, l2);
  int r3 = min(r1, r2);
  if (l3 > r3) return .0;
  return (double)(r3 - l3 + 1) / (r1 - l1 + 1);
}
void solve() {
  gen();
  int pl, pr, vl, vr, k;
  cin >> pl >> pr >> vl >> vr >> k;
  double ans = .0;
  if (pl <= luckys[int((luckys).size()) - 1]) {
    int l = pl;
    int i = 0;
    while (luckys[i] < l) ++i;
    while (l <= pr && int((luckys).size()) - i >= k) {
      int r = min(luckys[i], pr);
      int _l = luckys[i + k - 1];
      int _r = i + k < int((luckys).size()) ? luckys[i + k] - 1 : 1 << 30;
      ans += prob(pl, pr, l, r) * prob(vl, vr, _l, _r);
      l = r + 1;
      ++i;
    }
  }
  if (pr >= luckys[0]) {
    int r = pr;
    int i = int((luckys).size()) - 1;
    while (luckys[i] > r) --i;
    while (r >= pl && i + 1 >= k) {
      int l = max(luckys[i], pl);
      int _l = i - k >= 0 ? luckys[i - k] + 1 : 0;
      int _r = luckys[i - k + 1];
      ans += prob(pl, pr, l, r) * prob(vl, vr, _l, _r);
      if (k == 1)
        ans -= prob(pl, pr, luckys[i], luckys[i]) *
               prob(vl, vr, luckys[i], luckys[i]);
      r = l - 1;
      --i;
    }
  }
  printf("%.16lf\n", ans);
}
int main() {
  solve();
  return 0;
}