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 atexit
import io
import sys
 
buff = io.BytesIO()
sys.stdout = buff
 
 
@atexit.register
def write():
    sys.__stdout__.write(buff.getvalue())

def count(x,n):
	ans = 0
	if x&1:
		curr = x
		num = 1
		while curr <= n:
			ans += min(n-curr+1,num)
			num *= 2
			curr *= 2
		return ans

	curr = x
	num = 2
	while curr <= n:
		ans += min(n-curr+1,num)
		num *= 2
		curr *= 2
	return ans


n,k = map(int, raw_input().split())
ans_odd = 1
ans_even = 1

low = 1
high = n

while low <= high:
	mid = (low+high)/2
	if mid&1:
		mid += 1
	if count(mid,n) >= k:
		ans_even = mid
		low = mid+2
	else:
		high = mid-2

low = 1
high = n

while low <= high:
	mid = (low+high)/2
	if not (mid&1):
		mid += 1
	if count(mid,n) >= k:
		ans_odd = mid
		low = mid+2
	else:
		high = mid-2

print max(ans_odd, ans_even)




