ENIGMATH - PLAY WITH MATH


You would have been fed up with competitive programming questions so far, now it is time to solve little math.

Assume you have a equation A * x - B * y = 0

For a given value of A and B, find the minimum positive integer value of x and y that satisfies this equation.

Input

First line contains T, number of test cases 0 <= T <=1000 followed by T lines.

First line of each test case contains two space seperated integers A and B. 1 <= A, B <=1 000 000 000.

Output

For each test case, output a single line containing two integers x and y (seperated by a single space).

Example

Input:
1
2 3

Output:
3 2

Note:

  • Brute force won't pass the given constraint.
  • Negative number cases are avoided to make the problem easy.

hide comments
shikhargup_19: 2019-10-31 05:37:16


Last edit: 2019-10-31 05:51:22
sahruz_riyad: 2019-05-14 10:16:09

For large number (a*b or b*a) you have to find gcd..
in cpp use GCD = __gcd(a,b);
then a /= GCD and b/=GCD;
cout<<b<<' '<<a<<endl;

swap99: 2019-03-26 08:44:59

EASY PEASY

silentknight16: 2019-02-19 18:00:07

AC in one go!!

rock7897: 2019-01-19 20:18:01

Last edit: 2019-01-19 20:21:37
ankitpriyarup: 2018-12-19 19:25:17

Took me 2 WA finally ;)

dhia01: 2018-09-14 03:56:52

Cari KPK nya habistu , a sama b nya di bagi sama KPK nya

aman9598: 2018-05-30 15:53:35

its my 50th

rishapverma100: 2018-01-23 15:40:03

just find the lcm then divide lcm by a and b...x and y will be respective values....AC in one go...

vkash: 2018-01-03 11:24:07

AC in one Go...!!


Added by:B.R.ARVIND
Date:2013-09-12
Time limit:1s
Source limit:50000B
Memory limit:1536MB
Cluster: Cube (Intel G860)
Languages:All except: ASM64