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.
from math import *
from collections import *
import sys
sys.setrecursionlimit(10**9)

def check(ch):
	if(ch > n): return 0
	if(ch %2 == 0):
		if(ch == n): return 1
		else: 
			st = 4
			ans = 2
	else:
		st = 2
		ans  = 1
	ch *= 2
	while(ch < n):
		if(ch > n + 1 - st):
			ans += n+1-ch
			break
		else:
			ans += st
		st *= 2
		if(ch > n - ch):
			break
		else:
			ch *= 2
	return ans


mod = 10**9 + 7
n,x = map(int,input().split())

l = 1
r = n
ans1 = 0
while(r >= l):
	mid = (r+l)//2
	mid *= 2
	if(check(mid) >= x):
		l = mid//2 + 1
		ans1 = mid
	else:
		r = mid//2 - 1
	#print(mid,check(mid))
l = 1
r = n
ans2 = 0
while(r >= l):
	mid = (r+l)//2
	mid = mid*2-1
	if(check(mid) >= x):
		l = (mid+1)//2 + 1
		ans2 = mid
	else:
		r = (mid+1)//2 - 1
	#print(mid,check(mid))
print(max(ans1,ans2))