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 sys import stdin, stdout
from collections import Counter, defaultdict
pr=stdout.write
raw_input  = stdin.readline
def ni():
    return int(raw_input())


def li():
    return list(map(int,raw_input().split()))


def pn(n):
    stdout.write(str(n)+'\n')


def pa(arr):
    pr(' '.join(map(str,arr))+'\n')

# fast read function for total integer input

def inp():
    # this function returns whole input of
    # space/line seperated integers 
    # Use Ctrl+D to flush stdin.
    return (map(int,stdin.read().split()))
    
range = xrange # not for python 3.0+

# main code
def fun(x,n):
    ans=0
    if x%2:
        p=x
        c=1
    else:
        p=x
        c=2
    while p<=n:
        ans+=(min(n,p+c-1)-p+1)
        p*=2
        c*=2
    return ans
n,k=li()
l=1
r=n-(n/2)
odd=1
while l<=r:
    mid=(l+r)/2
    if fun((2*mid)-1,n)>=k:
        odd=(2*mid)-1
        l=mid+1
    else:
        r=mid-1
l=1
r=(n/2)
even=0
while l<=r:
    mid=(l+r)/2
    if fun(2*mid,n)>=k:
        l=mid+1
        even=2*mid
    else:
        r=mid-1
pn(max(odd,even))
