CF912B.New Year's Eve

Since Grisha behaved well last year, at New Year's Eve he was visited by Ded Moroz who brought an enormous bag of gifts with him! The bag contains n sweet candies from the good ol' bakery, each labeled from 1 to n corresponding to its tastiness. No two candies have the same tastiness.

The choice of candies has a direct effect on Grisha's happiness. One can assume that he should take the tastiest ones — but no, the holiday magic turns things upside down. It is the xor-sum of tastinesses that matters, not the ordinary sum!

A xor-sum of a sequence of integers _a_1, _a_2, ..., a__m is defined as the bitwise XOR of all its elements: , here denotes the bitwise XOR operation; more about bitwise XOR can be found here.

Ded Moroz warned Grisha he has more houses to visit, so Grisha can take no more than k candies from the bag. Help Grisha determine the largest xor-sum (largest xor-sum means maximum happiness!) he can obtain.

由于格里沙去年表现良好，新年除夕夜，圣诞老人（Дед Мороз）前来拜访，并带来了一个装满礼物的巨大袋子！袋中装有 $ n $ 颗来自古老而优秀的面包房的甜味糖果，每颗糖果按其美味程度从 $ 1 $ 到 $ n $ 编号。任意两颗糖果的美味程度均不相同。

所选糖果的组合会直接影响格里沙的幸福感。人们或许会认为他应挑选最美味的那些糖果——但事实并非如此，节日魔法将一切颠倒过来：真正起决定作用的是美味程度的异或和（xor-sum），而非普通的算术和！

序列 $ a_1, a_2, \dots, a_m $ 的异或和定义为其中所有元素的按位异或（bitwise XOR）：
，其中 表示按位异或运算；关于按位异或的更多内容，请参见此处。

圣诞老人警告格里沙，他还要去其他人家拜访，因此格里沙最多只能从袋中取走 $ k $ 颗糖果。请帮助格里沙确定他所能获得的最大异或和（最大异或和意味着最大的幸福感！）。

The sole string contains two integers n and k (1 ≤ k ≤ n ≤ 1018).

单行输入包含两个整数 $n$ 和 $k$（$1 \le k \le n \le 10^{18}$）。

Output one number — the largest possible xor-sum.

输出一个数字——最大的可能异或和。

In the first sample case, one optimal answer is 1, 2 and 4, giving the xor-sum of 7.

In the second sample case, one can, for example, take all six candies and obtain the xor-sum of 7.

在第一个样例中，一个最优解是选择编号为 1、2 和 4 的糖果，其异或和为 $7$。

在第二个样例中，例如可以选择全部六颗糖果，得到异或和 $7$。