## 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 > 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<