At first, let's define function f(x) as follows: $$$ \begin{matrix} f(x) & = & \left\{ \begin{matrix} x/2 & \mbox{if } x  is even \\\ x - 1 & \mbox{otherwise } \end{matrix} \right. \end{matrix} $$$

We can see that if we choose some value v and will apply function f to it, then apply f to f(v), and so on, we'll eventually get 1. Let's write down all values we get in this process in a list and denote this list as path(v). For example, path(1) = [1], path(15) = [15, 14, 7, 6, 3, 2, 1], path(32) = [32, 16, 8, 4, 2, 1].

Let's write all lists path(x) for every x from 1 to n. The question is next: what is the maximum value y such that y is contained in at least k different lists path(x)?

Formally speaking, you need to find maximum y such that \left| \{ x ~|~ 1 ≤ x ≤ n, y ∈ path(x) \} \right| ≥ k.

Input

The first line contains two integers n and k (1 ≤ k ≤ n ≤ 10^{18}).

Output

Print the only integer — the maximum value that is contained in at least k paths.

Examples

Input


11 3


Output


5


Input


11 6


Output


4


Input


20 20


Output


1


Input


14 5


Output


6


Input


1000000 100


Output


31248

Note

In the first example, the answer is 5, since 5 occurs in path(5), path(10) and path(11).

In the second example, the answer is 4, since 4 occurs in path(4), path(5), path(8), path(9), path(10) and path(11).

In the third example n = k, so the answer is 1, since 1 is the only number occuring in all paths for integers from 1 to 20.
import os, sys, math
import collections

if os.path.exists('testing'):
	name = os.path.basename(__file__)
	if name.endswith('.py'):
		name = name[:-3]
	src = open(name + '.txt', encoding='utf8')
	input = src.readline

def solve():
	def count_rec(fr, to):
		if n < fr:
			return 0
		if fr <= n <= to:
			return n - fr + 1
		return to - fr + 1 + count_rec(fr * 2, to * 2 + 1)
	def count(x):
		if x & 1:
			return count_rec(x, x)
		return count_rec(x, x + 1)
	def binary_search(fr, to, o):
		if fr == to:
			return fr * 2 + o
		s = (fr + to) // 2
		v = count(s * 2 + 2 + o)
		if v >= k:
			return binary_search(s + 1, to, o)
		return binary_search(fr, s, o)
	v1 = binary_search(0, n // 2, 0)
	v2 = binary_search(0, n // 2, 1)
	return max(v1, v2)

def integers():
	return map(int, input().strip().split()) 

def array_of_integers(q):
	return [
		tuple(map(int, input().strip().split())) for _ in range(q)
	]

n, k = integers()
res = solve()
print(res)
