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.
def gg(n,lol):
	ans = 0
	cur = 1
	lol2 = lol
	while(2*lol+1<=n):
		cur *= 2
		ans += cur
		lol = 2*lol+1
		lol2 *= 2
	if lol2*2 <= n:
		ans += n-lol2*2+1	
	return ans

n,k = map(int,input().split())
low = 1
high = n//2
res = 1
while low <= high:
	mid = (low+high)//2
	if gg(n,mid) >= k:
		res = mid
		low = mid+1
	else:
		high = mid-1
if n == k:
	print(1)
elif(gg(n,res)-1-gg(n,res*2) >= k):
	print(res*2+1)
else:
	print(res*2)					
