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?

Input

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 $n$, $x$ ($0\le n, x \le 10^{18}$).

Output

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 $5 \cdot 10^{18}$.

Note

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 $-1$.