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
def cnt(x):
    c = 0
    for i in range(65):
        if ((int(x)<< i) >  n):
            break
        if (x & 1):
            c += min((int(1)<<i),(n-(x<<i)+1))
        else:
            c += min((int(1)<<(i+1)),(n-(x<<i)+1))
    return c

(n,k) = [int(x) for x in input().split()]
#n,k=map(int,input().split())
if (k == 1):
    print(n)
    sys.exit(0)

l = 1
r = n+1
while (r-l > 1):
    m = (r+l)//2
    if (cnt(m*2-1) >= k):
        l = m
    else:
        r = m
ans1 = 2*l-1
l = 0
r = n+1
while (r-l>1):
    m = (l+r)//2
    if (cnt(m*2) >= k):
        l = m
    else:
        r = m
ans2 = 2*l
ans = max(ans1,ans2)
print(ans)
sys.exit(0)
