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())

def solve(n, num):
    ans = 0
    keta_num = num.bit_length()
    keta_n = n.bit_length()
    keta_diff = keta_n - keta_num
    if keta_diff < 0:
        return ans
    ans += (1 << keta_diff) - 1
    if (n >> keta_diff) > num:
        ans += 1 << keta_diff
    elif (n >> keta_diff) == num:
        ans += n & (((1 << keta_diff) - 1))
        ans += 1
    return ans

ok = 0
ng = n + 10
while abs(ok - ng) > 1:
    mid = (ok + ng) // 2
    val = mid * 2 + 1
    if solve(n, val) >= k:
        ok = mid
    else:
        ng = mid
ans1 = ok*2 + 1


ok = 0
ng = n + 10
while abs(ok - ng) > 1:
    mid = (ok + ng) // 2
    val = mid * 2
    if solve(n, val) + solve(n, val + 1) >= k:
        ok = mid
    else:
        ng = mid
ans2 = ok*2

print(max(ans1, ans2))