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

def cnt(y, n):
    # print("in count")
    #print(y)
    if y <= 1:
        return n
    if y > n:
        return 0
    if y % 2 == 1:
        return 1 + cnt(2 * y, n)
    c = 0
    p = 1
    while p*y <= n:
        mx = min(n, p*y + 2*p - 1)
        c += mx - p*y + 1
        p *= 2
    return c

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

if k == 1:
    print(n)
    sys.exit(0)

l, h = 1, n // 2
while l < h:
    m = (l + h) // 2
    #print("l = " + str(l))
    #print("h = " + str(h))
    #print("m = " + str(m))
    #print("cnt = " + str(cnt(m, n)))

    if cnt(2 * m, n) < k:
        h = m
    else:
        l = m + 1
mx_even = 2 * l - 2
    
l, h = 1, n // 2
while l < h:
    m = (l + h) // 2
    #print("l = " + str(l))
    #print("h = " + str(h))
    #print("m = " + str(m))
    #print("cnt = " + str(cnt(m, n)))

    if cnt(2 * m + 1, n) < k:
        h = m
    else:
        l = m + 1
mx_odd = 2 * l - 1
#assert(cnt(mx_odd, n) >= k)
#assert(cnt(mx_even, n) >= k)

mx_heur = -1
i = 0
while i < 20 and n - i > 0:
    if cnt(n - i, n) >= k:
        mx_heur = n - i
        break
    i += 1

print(max(mx_even, max(mx_odd, mx_heur)))
