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;
void file() {}
void fast() {
  std::ios_base::sync_with_stdio(0);
  cin.tie(NULL);
}
const int N = 1e9;
vector<int> lucky;
void generate(long long num) {
  if (num > N) return;
  if (num) lucky.push_back(num);
  generate(num * 10 + 4);
  generate(num * 10 + 7);
}
vector<int> get_all(int l, int r) {
  vector<int> ret;
  for (auto& it : lucky)
    if (it >= l && it <= r) ret.push_back(it);
  return ret;
}
inline double len(int l, int r) { return r - l + 1; }
int cnt_btw(int l, int r) {
  if (l > r) swap(l, r);
  return upper_bound(lucky.begin(), lucky.end(), r) -
         upper_bound(lucky.begin(), lucky.end(), l - 1);
}
pair<int, int> intersect(pair<int, int> l, pair<int, int> r) {
  if (l.first > l.second || r.first > r.second) return {1, -1};
  int x = max(l.first, r.first);
  int y = min(l.second, r.second);
  return {x, y};
}
int len_intersect(pair<int, int> seg) {
  int l = seg.first;
  int r = seg.second;
  if (l > r) return 0;
  return r - l + 1;
}
int get_prev(int num) {
  auto it = lower_bound(lucky.begin(), lucky.end(), num);
  if (it == lucky.begin()) return 1;
  it--;
  return *it + 1;
}
int get_next(int num) {
  auto it = upper_bound(lucky.begin(), lucky.end(), num);
  if (it == lucky.end()) return 1e9;
  return *it - 1;
}
int main() {
  file();
  fast();
  generate(0);
  sort(lucky.begin(), lucky.end());
  int pl, pr, vl, vr, k;
  cin >> pl >> pr >> vl >> vr >> k;
  int l = min(pl, vl);
  int r = max(pr, vr);
  vector<int> v1 = get_all(l, r);
  vector<pair<int, int> > arr;
  for (int i = k - 1; i < (int)(v1.size()); i++)
    arr.push_back({v1[i - k + 1], v1[i]});
  long long cnt = 0;
  for (auto& it : arr) {
    int prev = get_prev(it.first);
    int nxt = get_next(it.second);
    cnt += len_intersect(intersect({pl, pr}, {prev, it.first})) * 1LL *
           len_intersect(intersect({vl, vr}, {it.second, nxt}));
    pair<int, int> seg1 = intersect(intersect({pl, pr}, {prev, it.first}),
                                    intersect({vl, vr}, {it.second, nxt}));
    cnt += len_intersect(intersect({vl, vr}, {prev, it.first})) * 1LL *
           len_intersect(intersect({pl, pr}, {it.second, nxt}));
    pair<int, int> seg2 = intersect(intersect({vl, vr}, {prev, it.first}),
                                    intersect({pl, pr}, {it.second, nxt}));
    cnt -= len_intersect(intersect(seg1, seg2));
  }
  cout << fixed << setprecision(12);
  cout << cnt / (len(pl, pr) * len(vl, vr)) << '\n';
  return 0;
}