UVa 11388----GCD LCM

题目给出gcd(a,b)和lcm(a,b)，求a，b。

值得注意的是若结果有多组，输出a最小的那一组。

Description

 

IIU C   ONLINE   C ON TEST  2 008

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 integersa and b, a ≤ b, which has a GCD ofG and LCM of L. In case there is more than one pair satisfying the condition, output the pair for whicha 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

  

#include<stdio.h>int main(void){   int T,g,l;   scanf("%d",&T);   while(T--)   {      scanf("%d%d",&g,&l);  if(l%g!=0) printf("-1\n");  else   {     printf("%d %d\n",g,l);  }   }   return 0;}