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 separated 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 (separated 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.

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

hide comments
2018-09-14 03:56:52
Cari KPK nya habistu , a sama b nya di bagi sama KPK nya
2018-05-30 15:53:35
its my 50th
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...
2018-01-03 11:24:07
AC in one Go...!!
2017-12-09 18:50:13
dont read comments....
2017-12-02 03:23:07
Print minimum values of (x,y) not value of x&y, and not minimum value of x followed by matching y. Print each result on a separate line, don't insert additional newlines like comments suggest. Crap statement and crap judge: if a < b print the bigger value first else WA. WTF!
2017-09-05 05:15:37
Easiest question... Don't forget to print new line. Costed me one WA
2017-08-20 16:39:14
i used
a/=gcd(a,b);
b/=gcd(a,b);
n resulted 1 WA..:(
2017-07-30 03:45:42
AC in one go :)
2017-04-08 09:28:06
easy !! accepted in one go!!!use gcd
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