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 collections import Counter, defaultdict, deque
import bisect
from sys import stdin, stdout,setrecursionlimit
from itertools import repeat


# sys.stdin = open('input')
# n, k = map(int, raw_input().split())
# da = map(int, raw_input().split())
# db = map(int, raw_input().split())
def main():
    n, k = map(int, raw_input().split())
    def getk(num):
        total = 0
        if num%2==1:
            st, length = num, 1
        else:
            st, length = num, 2
        while st <= n:
            if st+length > n:
                total += max(0, n-st+1)
                break
            total += length
            st, length = st*2, length*2
        return total
    l, r = 1, n/2 + 1
    while l<r-1:
        mid = (l+r)/2
        if getk(mid*2) >=k:
            l = mid
        else:
            r = mid
    for i in range(l*2+1, l*2-2, -1):
        if getk(i) >= k:
            print i
            return







main()
