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

# inf = open('input.txt', 'r')
# reader = (map(int, line.split()) for line in inf)
reader = (map(int, line.split()) for line in sys.stdin)

def count(num):
    ans = 0
    if num % 2:
        add = 1
    else:
        add = 2
    while num + add - 1 <= n:
        ans += add
        add *= 2
        num *= 2
    if num <= n:
        ans += n - num + 1
    return ans

def func(n, k):
    if k == n:
        return 1
    # odd
    L = 0
    R = n // 2 + n % 2
    while L + 1 < R:
        m = (L + R) // 2
        num = 2 * m + 1
#         print(num, count(num))
        if count(num) >= k:
            L = m
        else:
            R = m
    m_odd = 2 * L + 1
    
    # even
    L = 1
    R = n // 2 + 1
    while L + 1 < R:
        m = (L + R) // 2
        num = 2 * m
#         print(num, count(num))
        if count(num) >= k:
            L = m
        else:
            R = m
    m_even = 2 * L
    return max(m_odd, m_even)
    
n, k = next(reader)
ans = func(n, k)
print(ans)

# inf.close()
