## 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.

 < Previous 1 2 3 4 Next > raghav9352: 2020-09-02 13:52:30 AC IN ONE GO!! manish_thakur: 2020-03-18 11:18:45 Don't look for hints in comments, observe for some time , and you'll be good to go! saketvajp_123: 2019-12-20 11:05:58 take input as 130 and 117 and then check .... for two different output(117 & 130 || 9 & 10) it accepts the answer. one we get using sqrt function and other by simple gcd. 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<