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 math

a,b = input().split()
a = int(a)
b = int(b)

if b==a:
    print("1")
else:
    ans=2
    start=3
    end=a
    while start<=end:
        mid=math.floor((start+end)>>1);
        #print(start,end,mid);
        if mid%2==0:
            gadha=math.ceil(math.log(math.ceil(b/2.0)+1,2))-1
            len_gadha=mid-2
            total_len=int(len_gadha*((1<<gadha)-1))
           # print(total_len,len_gadha,gadha,mid)
            if total_len+b+mid-1<=a:
                start=mid+1
                ans=mid
            else:
               # print("here")
                end=mid-1
        else:
            gadha=math.ceil(math.log(b+1,2))-1;
            len_gadha=mid-1
            total_len=int(len_gadha*((1<<gadha)-1))
           # print(total_len,len_gadha,gadha,mid)
            if total_len+b+mid-1<=a:
                start=mid+1
                ans=mid
            else:
                tmp=mid+1
                gadha=math.ceil(math.log(math.ceil(b/2.0)+1,2))-1
                len_gadha=tmp-2
                total_len=int(len_gadha*((1<<gadha)-1))
               # print(total_len,len_gadha,gadha,tmp)
                if total_len+b+tmp-1<=a:
                    start=tmp+1
                    ans=tmp
                else:
                    end=mid-1
    print(ans)
