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;
long long p1, p2, v1, v2, k, MAX = 1000000000;
vector<long long> a;
double ans;
void get(long long cur) {
  if (cur > MAX) {
    return;
  }
  a.push_back(cur);
  get(cur * 10 + 4), get(cur * 10 + 7);
}
double cnt(long long s, long long e, long long ind) {
  if (ind) {
    if (s > v2 || v1 > e) {
      return 0;
    }
    return (min(v2, e) - max(v1, s) + 1) / double(v2 - v1 + 1);
  } else {
    if (s > p2 || p1 > e) {
      return 0;
    }
    return (min(p2, e) - max(p1, s) + 1) / double(p2 - p1 + 1);
  }
}
void add(long long s, long long e) {
  long long s1 = 0, e1 = MAX;
  if (s) {
    s1 = a[s - 1] + 1;
  }
  if (e < a.size() - 1) {
    e1 = a[e + 1] - 1;
  }
  s = a[s], e = a[e];
  ans += cnt(s1, s, 0) * cnt(e, e1, 1), ans += cnt(s1, s, 1) * cnt(e, e1, 0);
  if (s == e && s >= p1 && s <= p2 && s >= v1 && s <= v2) {
    ans -= 1.0 / ((p2 - p1 + 1) * (v2 - v1 + 1));
  }
}
void solve() {
  cin >> p1 >> p2 >> v1 >> v2 >> k;
  get(4), get(7);
  sort(a.begin(), a.end());
  for (long long i = 0; i + k <= a.size(); ++i) {
    add(i, i + k - 1);
  }
  cout.precision(9);
  cout << fixed << ans << '\n';
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  solve();
  return 0;
}