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.
n,k=map(int,input().split())
def calc(z):
    global n
    if z>n:
        return False
    if z==0:
        return True
    if z%2==1:
        ans=1
        now=2*z
        co=1
        if k==1:
            return True
        elif now>n:
            return False
    else:
        ans=0
        now=z
        co=1
    #print(ans,now,co)
    while True:
        #print(n,now+2*co-1,now)
        #print(n-(now)+1)
        if n<=now+2*co-1:
            ans+=max(0,n-(now)+1)
            #print(ans)
            break
        else:
            now*=2
            co*=2
            ans+=(co)
            #print(ans)
    #print(ans)
    return ans>=k
realans=[]
#odd(2x-1)
l=1
r=n//2+2
while l+1<r:
    y=l+(r-l)//2
    if calc(2*y-1):
        l=y
    else:
        r=y
realans.append(2*l-1)
#even(2x)
l=0
r=n//2+2
while l+1<r:
    y=l+(r-l)//2
    if calc(2*y):
        l=y
    else:
        r=y
realans.append(2*l)
print(max(realans))