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())

left = 0
right = 10**18 + 2

#print("started")

p = [2**i for i in range(61)]

while right - left > 2:
    #print(left, right)
    mid = (right + left) // 2
    mid -= mid % 2
    l = 0
    r = 60
    if mid > n:
        right = mid
        continue
    while r - l > 1:
        m = (r + l) // 2
        if mid * p[m] <= n:
            l = m
        else:
            r = m
    cur = 0
    cur += p[l + 1] - 2 + min(n - p[l] * mid + 1, p[l + 1])
    if cur < k:
        right = mid
    else:
        left = mid

x = left

#print("finished 1")

left = 1
right = 10 ** 18 + 1

while right - left > 2:
    #print(left, right)
    mid = (right + left) // 2
    if mid % 2 == 0:
        mid += 1
    l = 0
    r = 60
    if mid > n:
        right = mid
        continue
    while r - l > 1:
        m = (r + l) // 2
        if mid * p[m] <= n:
            l = m
        else:
            r = m
    cur = 0
    cur += min(n - p[l] * mid + 1, p[l]) + p[l] - 1
    if cur < k:
        right = mid
    else:
        left = mid

#print("finished 2")

print(max(left, x))
