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 check(md, n, k):
    s=0
    
    temp, i = md, 1
    while temp<=n:
        s    += min(n-temp+1, i)
        temp *= 2
        i    *= 2
        
    if md%2==0:
        temp, i=md+1, 1
        while temp<=n:
            s    += min(n-temp+1, i)
            temp *= 2
            i    *= 2
        
    if s>=k:
        return True
    return False

n, k = map(int, input().split())

l, r = 0, n // 2 + 1

while r-l>1:
    md=(l+r) // 2 
    
    
    if check(2*md, n, k)==True:
        l=md
    else:
        r=md
        
even=2*l        
        
l, r = 0, n // 2 + 1

while r-l>1:
    md=(l+r) // 2 
    
    
    if check(2*md+1, n, k)==True:
        l=md
    else:
        r=md

odd=2*l+1

print(max(even, odd))        