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())
if (n == k):
    print(1)
    exit(0)
#odd

def calc(x):
    ans = 1
    
    if (x % 2 == 1):
        t = x
        while (True):
            x *= 2
            t = 2*t+1
            if (x > n):break
            ans += min(t, n) - x + 1
    else:
        if (x+1 <= n):
            t = x+1
            ans += 1
            while (True):
                x *= 2
                t = 2 * t + 1
                if (x > n): break
                ans += min(t, n) - x + 1
    return ans
l = 1
ans = 0
if (n%2 == 0):
    r = n-1
else:
    r = n
while (True):
    if (l == r):
        ans = l
        break
    if (l == r-1):
        if (l % 2 == 1):
            ans = l
        else:
            ans = r
        break

    m = (l+r) // 2
    if (m % 2 == 0): m += 1
    tt = calc(m)
    if (tt >= k):
        l = m
    else:
        r = m-1

l = 2
if (n%2 == 0):
    r = n
else:
    r = n-1

while (True):
    if (l == r):
        anss = l
        break
    if (l == r-1):
        if (l % 2 == 0):
            anss = l
        else:
            anss = r
        break
    m = (l+r) // 2
    if (m % 2 == 1): m += 1
    tt = calc(m)
    if (tt >= k):
        l = m
    else:
        r = m-1
print(max(anss, ans))

