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 int INF = 0x3f3f3f3f;
const double eps = 1e-9;
long long pl, pr, vl, vr, k;
long long l, r;
set<int> dp;
int rem[(int)1e5], sz;
void dfs(long long t) {
  if (t > r) return;
  if (l <= t) dp.insert(t);
  dfs(t * 10 + 4);
  dfs(t * 10 + 7);
}
void init() {
  cin >> pl >> pr >> vl >> vr >> k;
  l = min(pl, vl);
  r = max(pr, vr);
  dfs(0);
  sz = 0;
  for (set<int>::iterator it = dp.begin(); it != dp.end(); ++it)
    rem[sz++] = *it;
}
pair<int, int> intersect(int l0, int r0, int l1, int r1) {
  int l = max(l0, l1), r = min(r0, r1);
  if (l <= r) return make_pair(l, r);
  return make_pair(-1, -1);
}
struct Answer {
  pair<int, int> a, b;
  Answer() {}
  Answer(const pair<int, int> &a, const pair<int, int> &b) : a(a), b(b) {}
  bool operator<(const Answer &t) const {
    if (a != t.a) return a < t.a;
    if (b != t.b) return b < t.b;
    return false;
  }
};
set<Answer> final;
double solve() {
  pair<int, int> a, b;
  long long tot = (pr - pl + 1) * (vr - vl + 1);
  long long key = 0;
  for (int i = 0; i + k <= sz; ++i) {
    int j = i + k - 1;
    for (int x = 0; x < 2; ++x) {
      if (!i) {
        a = intersect(pl, pr, l, rem[i]);
      } else {
        a = intersect(pl, pr, rem[i - 1] + 1, rem[i]);
      }
      if (j + 1 == sz) {
        b = intersect(vl, vr, rem[j], r);
      } else {
        b = intersect(vl, vr, rem[j], rem[j + 1] - 1);
      }
      if (a.first != -1 && b.first != -1) {
        a.second--;
        b.first++;
        key += a.second - a.first + 1;
        key += b.second - b.first + 1;
        ++key;
        if (!x)
          final.insert(Answer(a, b));
        else
          final.insert(Answer(b, a));
      }
      swap(vl, pl), swap(vr, pr);
    }
  }
  if (k == 1) {
    int p0, p1;
    p0 = lower_bound(rem, rem + sz, pl) - rem;
    p1 = upper_bound(rem, rem + sz, pr) - rem;
    a = make_pair(p0, p1);
    p0 = lower_bound(rem, rem + sz, vl) - rem;
    p1 = upper_bound(rem, rem + sz, vr) - rem;
    b = make_pair(p0, p1);
    pair<int, int> del = intersect(a.first, a.second, b.first, b.second);
    if (del.first != -1) key -= del.second - del.first;
  }
  for (set<Answer>::iterator it = final.begin(); it != final.end(); ++it) {
    key += (long long)(it->a.second - it->a.first + 1) *
           (long long)(it->b.second - it->b.first + 1);
  }
  return (double)key / tot;
}
int main() {
  init();
  printf("%.10lf\n", solve());
  return 0;
}