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.
n,k=map(int,input().split())
j=n
odd=3
even=6
if j%2==0:
    n1=1
    n2=1

elif j%2!=0:
    n1=1
    n2=2

preodd=1
preeven=2
if k==n:
    print(1)
elif k==1:
    print(j)
elif k==2:
    if j%2==0:
        print(j-2)
    else:
        print(j-1)
else:
    pre=n
    j=n//2
    while (j >= 1):
        if j % 2 == 0:
            if pre % 2 == 0:
                if (n1 + preodd+1) >= k:
                    ans = j
                    break
                elif odd >= k:
                    ans = j - 1
                    break
                elif even >= k and j > 4:
                    ans = j - 2
                    break
                else:
                    n1 = (n1 + preodd + 1)
                    n2 = odd


            else:
                if (n2 + preodd+1) >= k:
                    ans = j
                    break
                elif odd >= k:
                    ans = j - 1
                    break
                elif even >= k and j > 4:
                    ans = j - 2
                    break
                else:
                    n1 = (n2 + preodd + 1)
                    n2 = odd


        else:
            if pre % 2 != 0:
                if (n2 + 1) >= k:
                    ans = j
                    break
                elif (preeven+n2+2) >= k:
                    ans = j - 1
                    break
                elif odd >= k and j > 4:
                    ans = j - 2
                    break
                elif even>=k and j>5:
                    ans=j-3
                    break
                else:
                    n1 = (n2 + 1)
                    n2 = preeven+n2+2

            else:
                if (n1 + 1) >= k:
                    ans = j
                    break
                elif (n1+preeven+3) >= k:
                    ans = j - 1
                    break
                elif odd >= k and j > 5:
                    ans = j - 2
                    break
                elif even>=k and j>4:
                    ans=j-3
                    break
                else:
                    n1 = (n1 + 1)
                    n2 = n1+preeven+2
        preodd = odd
        preeven = even
        odd = even + 1
        even = even + odd + 1
        pre = j
        j=j//2





    print(ans)




