Given positive integers $$$l$$$ and $$$x$$$, help SnowballSH find two integers $$$a$$$ and $$$b$$$ such that:
$$$^{\text{∗}}$$$The bitwise XOR of non-negative integers $$$x$$$, $$$y$$$, $$$x \oplus y$$$, is defined as follows:
When $$$x \oplus y$$$ is written in binary, the digit in the $$$2^k$$$ ($$$k \ge 0$$$) place is $$$1$$$ if exactly one of $$$x,y$$$ has $$$1$$$ in the $$$2^k$$$ place when written in binary, and $$$0$$$ otherwise.
For example, $$$3 \oplus 5 = 6$$$ (in binary representation: $$$011 \oplus 101=110$$$).
$$$^{\text{†}}$$$$$$\mathrm{popcount}(x)$$$ represents the number of $$$1$$$s in the binary representation of $$$x$$$.
For example, $$$13=1101_{(2)}$$$, so $$$\mathrm{popcount}(13)=3$$$.
The first line contains a single integer $$$t$$$ $$$(1 \le t \le 10^4)$$$ — the number of test cases.
The only line of each test case contains two integers $$$l$$$ and $$$x$$$ $$$(1 \le l \le 10^{17}, 1 \le x \le 59)$$$.
Output two integers $$$a$$$ and $$$b$$$ such that $$$l \le a \le b \le 10^{18}$$$ and $$$\mathrm{popcount}(a \oplus b) = x$$$.
If there are multiple answers, you may output any of them.
It can be shown that an answer always exists.
31 12 215 3
6 7 2 22 20 25
For the first test case, we have $$$l=1$$$ and $$$x=1$$$. $$$a=6$$$ and $$$b=7$$$ works, because $$$1 \le 6 \le 7 \le 10^{18}$$$ and $$$\mathrm{popcount}(6 \oplus 7) = \mathrm{popcount}(1) = 1$$$.
For the second test case, we have $$$l=2$$$ and $$$x=2$$$. $$$a=2$$$ and $$$b=22$$$ works, because $$$1 \le 2 \le 22 \le 10^{18}$$$ and $$$\mathrm{popcount}(2 \oplus 22) = \mathrm{popcount}(20) = \mathrm{popcount}(10100_{(2)}) = 2$$$.
For the third test case, we have $$$l=15$$$ and $$$x=3$$$. $$$a=20$$$ and $$$b=25$$$ works, because $$$15 \le 20 \le 25 \le 10^{18}$$$ and $$$\mathrm{popcount}(20 \oplus 25) = \mathrm{popcount}(13) = \mathrm{popcount}(1101_{(2)}) = 3$$$.
