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>
long long bigmod(long long a, long long b) {
  if (b == 0) return 1;
  long long ret = bigmod(a, b / 2);
  return ret * ret % 1000000007 * (b % 2 ? a : 1) % 1000000007;
}
using namespace std;
double a, b, c, d;
int k;
long long tap(long long x, long long y, long long x2, long long y2) {
  if (y2 < x || x2 > y) return 0;
  return min(y, y2) - max(x, x2) + 1;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> a >> b >> c >> d >> k;
  vector<long long> L, last;
  L.push_back(0);
  last.push_back(0);
  for (int i = 1; i <= 10; i++) {
    vector<long long> cur;
    for (auto h : last) cur.push_back(h * 10 + 4);
    for (auto h : last) cur.push_back(h * 10 + 7);
    for (auto h : cur) L.push_back(h);
    swap(last, cur);
  }
  sort(L.begin(), L.end());
  L.push_back(2e15);
  double ans = 0;
  for (long long i = k; i < L.size() - 1; i++) {
    long long l = L[i - k] + 1, r = L[i - k + 1];
    long long l1 = L[i], r1 = L[i + 1] - 1;
    if (k == 1 && b >= L[i] && c <= L[i]) ans--;
    ans += tap(l, r, a, b) * tap(l1, r1, c, d) +
           tap(l, r, c, d) * tap(l1, r1, a, b);
  }
  ans /= (b - a + 1) * (d - c + 1);
  printf("%.12lf", ans);
}