#P1584A. Mathematical Addition
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ID: 6116
Type: RemoteJudge
1000ms
256MiB
Tried: 0
Accepted: 0
Difficulty: (None)
Uploaded By:
Hydro
Tags> Mathematical Addition
No submission language available for this problem.
Description
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.
Input
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.
Output
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.
Samples
4
1 1
2 3
3 5
6 9
-1 1
-4 9
-18 50
-4 9
Note
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: $\frac{-4}{6} + \frac{9}{9} = \frac{1}{3} = \frac{-4 + 9}{6 + 9}$.