codeforces#P1584A. Mathematical Addition

Mathematical Addition

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 uu and vv, find any pair of integers (not necessarily positive) xx, yy, such that: xu+yv=x+yu+v.\frac{x}{u} + \frac{y}{v} = \frac{x + y}{u + v}.
  • The solution x=0x = 0, y=0y = 0 is forbidden, so you should find any solution with (x,y)(0,0)(x, y) \neq (0, 0).

Please help Ivan to solve some equations of this form.

The first line contains a single integer tt (1t1031 \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 uu and vv (1u,v1091 \leq u, v \leq 10^9) — the parameters of the equation.

For each test case print two integers xx, yy — a possible solution to the equation. It should be satisfied that 1018x,y1018-10^{18} \leq x, y \leq 10^{18} and (x,y)(0,0)(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 tt (1t1031 \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 uu and vv (1u,v1091 \leq u, v \leq 10^9) — the parameters of the equation.

Output

For each test case print two integers xx, yy — a possible solution to the equation. It should be satisfied that 1018x,y1018-10^{18} \leq x, y \leq 10^{18} and (x,y)(0,0)(x, y) \neq (0, 0).

We can show that an answer always exists. If there are multiple possible solutions you can print any.

Samples

输入数据 1

4
1 1
2 3
3 5
6 9

输出数据 1

-1 1
-4 9
-18 50
-4 9

Note

In the first test case: 11+11=0=1+11+1\frac{-1}{1} + \frac{1}{1} = 0 = \frac{-1 + 1}{1 + 1}.

In the second test case: 42+93=1=4+92+3\frac{-4}{2} + \frac{9}{3} = 1 = \frac{-4 + 9}{2 + 3}.

In the third test case: 183+505=4=18+503+5\frac{-18}{3} + \frac{50}{5} = 4 = \frac{-18 + 50}{3 + 5}.

In the fourth test case: 46+99=13=4+96+9\frac{-4}{6} + \frac{9}{9} = \frac{1}{3} = \frac{-4 + 9}{6 + 9}.