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 binary_search(lb, ub, criterion):
    ok = lb
    ng = ub
    while ng - ok > 1:
        mid = (ng + ok) // 2
        if criterion(mid):
            ok = mid
        else:
            ng = mid
    return ok


n, k = list(map(int, input().split()))
odd_acc_count = 0
odd_count = 1
depth = 0
while odd_acc_count < k:
    odd_acc_count += odd_count
    odd_count *= 2
    depth += 1

residual = k - (odd_acc_count - odd_count // 2) - 1
odd_max_idx = binary_search(0, n + 2, lambda x:
                            (2 * x + 1) * (1 << (depth - 1)) + residual <= n)
ans_odd = 2 * odd_max_idx + 1

even_acc_count = 0
even_count = 2
depth = 0
while even_acc_count < k:
    even_acc_count += even_count
    even_count *= 2
    depth += 1
residual = k - (even_acc_count - even_count // 2) - 1
even_max_idx = binary_search(0, n + 2, lambda x:
                             2 * x * (1 << (depth - 1)) + residual <= n
                             )

ans_even = 2 * even_max_idx
print(max(ans_odd, ans_even))
