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>
#pragma comment(linker, "/STACK:200000000")
using namespace std;
const int INF = (int)1E9;
const long long INF64 = (long long)1E18;
const double EPS = 1E-9;
const double PI = acos((double)0) * 2;
vector<int> num;
int l1, r1, l2, r2, k;
bool read() { return !!(cin >> l1 >> r1 >> l2 >> r2 >> k); }
long long prob(int first1, int last1, int first2, int last2) {
  first1 = max(first1, l1);
  last1 = min(last1, r1);
  first2 = max(first2, l2);
  last2 = min(last2, r2);
  if (first1 > last1 || first2 > last2) return 0;
  return (last1 - first1 + 1) * 1ll * (last2 - first2 + 1);
}
void solve() {
  long long ans = 0;
  for (int i = 0; i < int(num.size() - k + 1); ++i) {
    int first = num[i], last = num[i + k - 1], pfirst = i ? num[i - 1] + 1 : 1,
        nlast = i + k < (int)num.size() ? num[i + k] - 1 : INF;
    ans += prob(pfirst, first, last, nlast);
    ans += prob(last, nlast, pfirst, first);
    if (k == 1) ans -= prob(first, first, first, first);
  }
  double dans = ans * 1.0 / ((r1 - l1 + 1) * 1ll * (r2 - l2 + 1));
  printf("%.20lf\n", dans);
}
int main() {
  vector<int> prev;
  prev.push_back(0);
  for (;;) {
    vector<int> cur;
    for (int i = 0; i < int(prev.size()); ++i) {
      long long n = prev[i] * 10ll;
      if (n + 4 <= INF) cur.push_back(int(n + 4));
      if (n + 7 <= INF) cur.push_back(int(n + 7));
    }
    if (cur.empty()) break;
    for (int i = 0; i < int(cur.size()); ++i) num.push_back(cur[i]);
    prev = cur;
  }
  if (!read()) throw;
  solve();
}