1271_E. Common Number
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.
```python
n, k = map(int,input().split())
p = 1 
w = 0
while p * 2 <= k:
	p *= 2
	w += 1
kk = k - p
#2**w * tr <= n - kk
odd = (n-kk)//2**w
if odd % 2 == 0:
	odd -= 1

p = 1
w = 0
while 2 * (2*p - 1) < k:
	p *= 2
	w += 1
poz = k - 2*(p-1)
if poz == 0:
	#2**(w-1) * strzal + (2**w-1) <= n
	even = (n-(2**w-1))//2**(w-1)
	if even % 2 == 1:
		even -= 1
else:
	if poz <= p*4343:
		#porownanie z zerem na koncu
		strzal = (n+1-poz)//p
		if strzal % 2 == 1:
			strzal -= 1
	even = strzal
print(max(even,odd))

```