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 _calc(prefix, n):
    if prefix > n:
        return 0
    prefix = bin(prefix)[2:]
    n = bin(n)[2:]
    plen = len(prefix)
    nlen = len(n)

    res = (1 << (nlen - plen)) - 1
    nprefix = n[:plen]
    if nprefix == prefix:
        for i in range(plen, nlen):
            if n[i] == '1':
                res += 1 << (nlen - i - 1)
        res += 1  # n self
    elif nprefix > prefix:
        res += 1 << (nlen - plen)
    return res


def calc(prefix, n):
    if prefix % 2 == 0:
        return _calc(prefix, n) + _calc(prefix + 1, n)
    return _calc(prefix, n)


n, k = map(int, input().split())
if n == k:
    print(1)
    exit(0)

ans = 1
# even
l, r = 1, n >> 1
while l < r:
    mid = (l + r + 1) >> 1
    tot = calc(mid * 2, n)
    if tot >= k:
        l = mid
    else:
        r = mid - 1
l *= 2
if calc(l, n) >= k:
    ans = max(ans, l)

# odd
l, r = 0, (n - 1) >> 1
while l < r:
    mid = (l + r + 1) >> 1
    tot = calc(mid * 2 + 1, n)
    if tot >= k:
        l = mid
    else:
        r = mid - 1
l = l * 2 + 1
if calc(l, n) >= k:
    ans = max(ans, l)

print(ans)
