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.
'''input
1000000 100
'''

import sys
from collections import Counter
debug = 0
readln = sys.stdin.readline
#sys.setrecursionlimit(1000000)

def write(s):
    sys.stdout.write(str(s))

def writeln(s):
    sys.stdout.write(str(s))
    sys.stdout.write('\n')

def readint():
    return int(readln().strip())

def readints():
    return map(int, readln().split())

def readstr():
    return readln().strip()

def readstrs():
    return readln().split()

def dprint(*args):
    if debug: print(' '.join(map(str, args)))

def mcnt(b, n):
    ans = 0
    i = 1
    l, r = b, b
    while True:
        dprint("mcnt: %d %d %d" % (l, r, n))
        if r % 2 == 0: r += 1

        if n < l:
            return ans
        if n < r:
            return ans + (n - l)+1
        ans += r - l + 1
        l = l * 2
        r = r * 2

def solve(n, k):
    l = 1
    r = n
    ans = 0
    while l <= r:
        mid1 = (l + r) / 2
        mid2 = mid1 + 1
        cnt1 = mcnt(mid1, n)
        dprint(mid1, n, cnt1)
        cnt2 = mcnt(mid2, n)
        dprint(mid2, n, cnt2)
        if cnt2 >= k:
            ans = max(ans, mid2)
            l = mid1 + 1
        elif cnt1 >= k:
            ans = max(ans, mid1)
            l = mid1 + 1
        else:
            r = mid1 - 1
    return ans

n,k = readints()
writeln(solve(n,k))
