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.
from sys import stdin, stdout


def getmaxval(n, k):

    if n == k:
        return 1

    # odd binary search
    ol = 1
    #oh = n // 2 + n % 2
    oh = n // 2

    while ol < oh:
        # print('ol:' + str(ol) + '  oh:' + str(oh))

        om = (ol + oh + 1) // 2
        cnt = counteven(om * 2, n)

        #print('om:' + str(om * 2 ) + '  cnt:' + str(cnt))
        # breakpoint()

        if cnt >= k:
            ol = om
        else:
            oh = om - 1

    r1 = ol * 2
    if r1 < n:
        cnt2 = countodd(r1 + 1, n)
        if cnt2 >= k:
            return r1+1

    return r1


def counteven(b, n):
    res = 2
    inc = 2
    mv = b + 1

    #print('inc:' + str(inc) + ' mv:' + str(mv) + ' res:' + str(res))

    while mv <= n:
        inc *= 2
        res += inc
        mv = b * inc//2 + inc - 1

        #print('inc:' + str(inc) + ' mv:' + str(mv) + ' res:' + str(res))

    if (mv - inc) <= n < mv:
        res -= (mv - n)
    elif n < (mv - inc):
        res -= inc

    return res


def countodd(b, n):
    res = 1
    inc = 1
    mv = b

    while mv <= n:
        inc *= 2
        mv = b * inc + inc - 1
        res += inc

        #print('inc:' + str(inc) + ' mv:' + str(mv) + ' res:' + str(res))

    if (mv - inc) <= n < mv:
        res -= (mv - n)
    elif n < (mv - inc):
        res -= inc

    return res


if __name__ == '__main__':
    nk = list(map(int, stdin.readline().split()))
    n = nk[0]
    k = nk[1]

    #res = counteven(4, 11)
    #res = countodd(5, 11)
    #res = counteven(6, 11)

    #res = counteven(6, 23)
    #res = counteven(6, 24)
    #res = counteven(6, 25)
    #res = counteven(6, 26)
    #res = counteven(6, 27)
    #res = counteven(6, 28)

    res = getmaxval(n, k)
    stdout.write(str(res))
