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 sys
input=sys.stdin.readline
n,k=map(int,input().split())
# even
l=0;r=n
while r-l>1:
    mid=(l+r)//2
    x=mid*2
    cnt=0
    if x<=n:
        cnt+=1
    if x+1<=n:
        cnt+=1
    v=2
    while v*x<=n:
        if v*x+2*v-1<=n:
            cnt+=2*v
            v*=2
        else:
            cnt+=n-v*x+1
            break
    if cnt>=k and x<=n:
        l=mid
    else:
        r=mid
ans=l*2

l=1;r=n
while r-l>1:
    mid=(l+r)//2
    x=mid*2-1
    cnt=0
    if x<=n:
        cnt+=1
    v=2
    while v*x<=n:
        if v*x+v-1<=n:
            cnt+=v
            v*=2
        else:
            cnt+=n-v*x+1
            break
    if cnt>=k and x<=n:
        l=mid
    else:
        r=mid
ans=max(ans,2*l-1)
print(ans)