题目：

大意是这样的： 给出两个数gcd和lcm，求出满足条件的a和b，其中a尽可能的小。

I I U C   O N L I N E   C O N T E S T   2 0 0 8

Problem D: GCD LCM

Input: standard input
Output: standard output

 

The GCD of two positive integers is the largest integer that divides both the integers without any remainder. The LCM of two positive integers is the smallest positive integer that is divisible by both the integers. A positive integer can be the GCD of many pairs of numbers. Similarly, it can be the LCM of many pairs of numbers. In this problem, you will be given two positive integers. You have to output a pair of numbers whose GCD is the first number and LCM is the second number.

 

Input

The first line of input will consist of a positive integer T. T denotes the number of cases. Each of the next T lines will contain two positive integer, G and L.

 

Output

For each case of input, there will be one line of output. It will contain two positive integers a and b, a ≤ b, which has a GCD of G and LCM of L. In case there is more than one pair satisfying the condition, output the pair for which a is minimized. In case there is no such pair, output -1.

 

Constraints

-           T ≤ 100

-           Both G and L will be less than 231.

 

Sample Input

Output for Sample Input

2

1 2

3 4

1 2

-1

 

Problem setter: Shamim Hafiz

数学题一旦发现了规律代码通常很短，但是蛮干的话多半没有好下场。

(a*b)/gcd=lcm.  (a*b)/(gcd*gcd)*gcd=lcm-->(a*b)/(gcd*gcd)=lcm/gcd -->lcm/gcd=(a/gcd)*(b/gcd)
 so if exist a,b must be gcd, b must be lcm.
#include <iostream>
#include<cstdio> 
using namespace std;
typedef long long LL;
int main(int argc, char *argv[]) {
  //freopen("cin.txt","r",stdin);
  LL T,G,L;
  while(cin>>T){
    while(T--){
      LL a,b;
      scanf("%lld%lld",&G,&L);
      if(L%G==0)printf("%lld %lld\n",G,L);
      else printf("-1\n");
    }
  }
  return 0;
}