CF1775C.Interesting Sequence

Petya and his friend, robot Petya++, like to solve exciting math problems.

One day Petya++ came up with the numbers $n$ and $x$ and wrote the following equality on the board: $$n\ \&\ (n+1)\ \&\ \dots\ \&\ m = x,$$ where $\&$ denotes the bitwise AND operation. Then he suggested his friend Petya find such a minimal $m$ ($m \ge n$) that the equality on the board holds.

Unfortunately, Petya couldn't solve this problem in his head and decided to ask for computer help. He quickly wrote a program and found the answer.

Can you solve this difficult problem?

佩佳和他的朋友——机器人佩佳++，喜欢解决有趣的数学问题。

有一天，佩佳++想出了两个数 $n$ 和 $x$，并在黑板上写下了如下等式：

$$n\ \&\ (n+1)\ \&\ \dots\ \&\ m = x,$$

其中 $\&$ 表示按位与运算。接着，他建议他的朋友佩佳找出满足该等式的最小 $m$（要求 $m \ge n$）。

不幸的是，佩佳无法心算解决这个问题，于是决定求助于计算机。他迅速编写了一个程序，并找到了答案。

你能否解决这个难题？

Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 2000$). The description of the test cases follows.

The only line of each test case contains two integers $n$, $x$ ($0\le n, x \le 10^{18}$).

每个测试包含多个测试用例。第一行包含测试用例的数量 $t$（$1 \le t \le 2000$）。随后是各测试用例的描述。

每个测试用例仅有一行，包含两个整数 $n$、$x$（$0\le n, x \le 10^{18}$）。

For every test case, output the smallest possible value of $m$ such that equality holds.

If the equality does not hold for any $m$, print $-1$ instead.

We can show that if the required $m$ exists, it does not exceed $5 \cdot 10^{18}$.

对于每个测试用例，输出使得等式成立的最小可能的 $m$ 值。

如果不存在使等式成立的 $m$，则输出 $-1$。

可以证明：若所求的 $m$ 存在，则其值不超过 $5 \cdot 10^{18}$。

In the first example, $10\ \&\ 11 = 10$, but $10\ \&\ 11\ \&\ 12 = 8$, so the answer is $12$.

In the second example, $10 = 10$, so the answer is $10$.

In the third example, we can see that the required $m$ does not exist, so we have to print $-1$.

在第一个例子中，$10\ \&\ 11 = 10$，但 $10\ \&\ 11\ \&\ 12 = 8$，因此答案为 $12$。

在第二个例子中，$10 = 10$，因此答案为 $10$。

在第三个例子中，我们可以看出所要求的 $m$ 不存在，因此需输出 $-1$。