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<long long> luk;
inline long long intersect(pair<long long, long long> a,
                           pair<long long, long long> b) {
  long long res = min(a.second, b.second) - max(a.first, b.first) + 1;
  return max(0ll, res);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  luk.push_back(0);
  for (int i = 1; i <= 10; i += 1) {
    for (long long j = 0; j < (1LL << i); j++) {
      long long num = 0;
      for (int k = 0; k < i; k++) {
        num *= 10;
        num += ((1 << k) & j) ? 7 : 4;
      }
      luk.push_back(num);
    }
  }
  luk.push_back(1e18);
  sort((luk).begin(), (luk).end());
  pair<long long, long long> p, v;
  int k;
  cin >> p.first >> p.second >> v.first >> v.second >> k;
  long long res = 0;
  for (int i = 1; i <= ((int)(luk).size()) - k - 1; i += 1) {
    pair<long long, long long> st, en;
    st.first = luk[i - 1] + 1;
    st.second = luk[i];
    en.first = luk[i + k - 1];
    en.second = luk[i + k] - 1;
    res += intersect(st, v) * intersect(en, p);
    res += intersect(st, p) * intersect(en, v);
    if (k == 1 && intersect(v, make_pair((luk[i]), (luk[i]))) &&
        intersect(p, make_pair((luk[i]), (luk[i]))))
      --res;
  }
  printf("%.10lf\n",
         res / ((p.second - p.first + 1.0) * (v.second - v.first + 1.0)));
  return 0;
}