D = gets.to_i

def f(x, limit = 1000)
  len = 0
  cnt = 0

  while x > 0 && cnt < limit
    len += x
    x /= 2
    cnt += 1
    puts len if limit == 1000
  end

  len
end

ans = D

(1..100).each do |limit|
  ng = 0
  ok = 10 ** 19

  while (ok - ng).abs >= 2
    x = (ok + ng) / 2

    if D <= f(x, limit)
      ok = x
    else
      ng = x
    end
  end

  ans = ok if f(ok, limit) == D && ans > ok
end

puts ans