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 ans(m, n):
    q1, q2, k = m, 2, 0
    while q1 <= n:
        q1 *= 2
        k += q2
        q2 *= 2
    q2 //= 2
    q1 //= 2
    if n-q1 < q2:
        return k+n-q1-q2+1
    return k


n, k = map(int, input().split())
if k == n:
    print(1)
elif k == 1:
    print(n)
else:
    l, r = 1, n//2+1
    while r-l > 1:
        m = (l+r)//2
        if ans(2*m, n) >= k:
            l = m
        else:
            r = m
    l1, r1 = 1, n // 2 + 1
    while r1 - l1 > 1:
        m = (l1 + r1) // 2
        if ans(2 * m, n) >= k-1:
            l1 = m
        else:
            r1 = m
    print(max(l1, 2*l))
