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;
int N, c;
long double l, r, a, b;
long double A[100100];
long double sol, lenl, lena;
void DFS(long long n) {
  if (n > 1e9) return;
  A[++N] = n;
  DFS(n * 10 + 4);
  DFS(n * 10 + 7);
}
long double prob(long double lol, long double hil, long double loa,
                 long double hia) {
  lol = max(lol, l);
  hil = min(hil, r);
  loa = max(loa, a);
  hia = min(hia, b);
  long double cl = (hil - lol + 1);
  long double ca = (hia - loa + 1);
  if (min(cl, ca) <= 0) return 0;
  return (cl * lenl) * (ca * lena);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout << setprecision(12) << fixed;
  DFS(4);
  DFS(7);
  sort(A, A + N);
  A[0] = 0;
  A[N + 1] = 1100100100;
  cin >> l >> r >> a >> b >> c;
  lenl = (long double)(1) / (r - l + 1);
  lena = (long double)(1) / (b - a + 1);
  for (int I = c; I <= N; I++) {
    sol += prob(A[I - c] + 1, A[I - c + 1], A[I], A[I + 1] - 1);
    sol += prob(A[I], A[I + 1] - 1, A[I - c] + 1, A[I - c + 1]);
    if (c == 1) sol -= prob(A[I], A[I], A[I], A[I]);
  }
  cout << sol << endl;
}