CF1584A Mathematical Addition

Ivan decided to prepare for the test on solving integer equations. He noticed that all tasks in the test have the following form: - You are given two positive integers $ u $ and $ v $, find any pair of integers (**not necessarily positive**) $ x $, $ y $, such that: $$\frac{x}{u} + \frac{y}{v} = \frac{x + y}{u + v}.$$ - The solution $x = 0$, $ y = 0 $ is forbidden, so you should find any solution with $(x, y) \neq (0, 0)$. Please help Ivan to solve some equations of this form.

The first line contains a single integer $ t $ ( $ 1 \leq t \leq 10^3 $ ) — the number of test cases. The next lines contain descriptions of test cases. The only line of each test case contains two integers $ u $ and $ v $ ( $ 1 \leq u, v \leq 10^9 $ ) — the parameters of the equation.

For each test case print two integers $ x $ , $ y $ — a possible solution to the equation. It should be satisfied that $ -10^{18} \leq x, y \leq 10^{18} $ and $ (x, y) \neq (0, 0) $ . We can show that an answer always exists. If there are multiple possible solutions you can print any.

In the first test case: $ \frac{-1}{1} + \frac{1}{1} = 0 = \frac{-1 + 1}{1 + 1} $ . In the second test case: $ \frac{-4}{2} + \frac{9}{3} = 1 = \frac{-4 + 9}{2 + 3} $ . In the third test case: $ \frac{-18}{3} + \frac{50}{5} = 4 = \frac{-18 + 50}{3 + 5} $ . In the fourth test case: $ \frac{-4}{6} + \frac{9}{9} = \frac{1}{3} = \frac{-4 + 9}{6 + 9} $ .