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 paths(x):
    c = 1 if x % 2 else 2
    a = 0
    m = 1
    while m * x + c * m - 1 <= n:
        a += c * m
        m *= 2
    a += max(-1, n - m * x) + 1
    return a


def search_odd(k, i, j):
    if i == j:
        return 2 * i + 1
    if i + 1 == j:
        return 2 * j + 1 if paths(2 * j + 1) >= k else 2 * i + 1
    m = (i + j) >> 1
    if paths(2 * m + 1) >= k:
        return search_odd(k, m, j)
    else:
        return search_odd(k, i, m - 1)


def search_even(k, i, j):
    if i == j:
        return 2 * i
    if i + 1 == j:
        return 2 * j if paths(2 * j) >= k else 2 * i
    m = (i + j) >> 1
    if paths(2 * m) >= k:
        return search_even(k, m, j)
    else:
        return search_even(k, i, m - 1)


def search(k):
    return max(search_even(k, 0, n // 2), search_odd(k, 0, (n - 1) // 2))


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