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;
const int INF = (int)2E9;
const long long LINF = (long long)9E18;
const long double PI = acos(-1.0);
const long double EPS = 1E-11;
int pl, pr, vl, vr, k;
int A[1000000], N, L_;
long long Tu, Mau;
void try_(int i, int val) {
  if (i == L_ + 1) {
    N++;
    A[N] = val;
    return;
  }
  try_(i + 1, val * 10 + 4);
  try_(i + 1, val * 10 + 7);
}
long long Len(int a, int b, int x, int y) {
  if (b < x || y < a) return 0;
  return min(b, y) - max(a, x) + 1;
}
void call_1() {
  for (int i = (1); i <= (N); ++i)
    if (A[i] > pr) {
      if (i + k - 1 >= N)
        Tu +=
            Len(A[i - k - 1] + 1, A[i - k], vl, vr) * (long long)(pr - pl + 1);
      break;
    }
  for (int i = (N); i >= (1); --i)
    if (A[i] < pl) {
      if (i + k <= N)
        Tu +=
            Len(A[i + k], A[i + k + 1] - 1, vl, vr) * (long long)(pr - pl + 1);
      break;
    }
}
void call_2() {
  for (int i = (1); i <= (N - k + 1); ++i)
    Tu += Len(pl, pr, A[i - 1] + 1, A[i]) *
          Len(A[i + k - 1], A[i + k] - 1, vl, vr);
  for (int i = (k); i <= (N); ++i) {
    Tu += Len(pl, pr, A[i], A[i + 1] - 1) *
          Len(vl, vr, A[i - k] + 1, A[i - k + 1]);
    if (k == 1 && vl <= A[i] && A[i] <= vr && pl <= A[i] && A[i] <= pr) Tu--;
  }
}
int main() {
  cin >> pl >> pr >> vl >> vr >> k;
  for (int i = (1); i <= (9); ++i) {
    L_ = i;
    try_(1, 0);
  }
  Tu = 0;
  A[0] = 0;
  A[N + 1] = 1000000001;
  int ok = -1;
  for (int i = (1); i <= (N); ++i)
    if (pl <= A[i] && A[i] <= pr) {
      ok = 1;
      break;
    }
  if (ok == -1)
    call_2();
  else
    call_2();
  Mau = (long long)(pr - pl + 1) * (long long)(vr - vl + 1);
  printf("%.9lf", (double)Tu / Mau);
}