## 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 <= AB <=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.

## Code Examples

### #1 Code Example with C++ Programming

```Code - C++ Programming```

``````#include <bits/stdc++.h>
#include <climits>
using namespace std;
typedef long long ll;
typedef pair<int, int> ii;
typedef vector<ii> vii;
typedef vector<int> vi;

#define INF 1000000000
#define MOD (ll)1000000007
#define pb 	push_back
#define EPS 1e-9
#define FOR(i, n)	for(int i = 0;i  <  n; i++)

template <typename T> T gcd(T a, T b){return (b==0)?a:gcd(b,a%b);}
template <typename T> T lcm(T a, T b){return a*(b/gcd(a,b));}
template <typename T> T mod_exp(T b, T p, T m){T x = 1;while(p){if(p&1)x=(x*b)%m;b=(b*b)%m;p=p>>1;}return x;}
template <typename T> T invFermat(T a, T p){return exp(a, p-2, p);}
template <typename T> T exp(T b, T p){T x = 1;while(p){if(p&1)x=(x*b);b=(b*b);p=p>>1;}return x;}

int t;
int main(){
cin>>t;
while(t--){
ll a, b;
cin>>a>>b;
ll g = gcd(a, b);
ll l = a*(b/g);
cout<<l/a<<' '<<l/b<<endl;
}
return 0;
}``````
Copy The Code &

Input

cmd
1
2 3

Output

cmd
3 2