UVa 11388

Description

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, Gand 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 Gand 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 Lwill be less than (2^{31}).

Sample Input

2
1 2
3 4

Sample Output

1 2
-1

Resume

输入两个整数G、L，求两个数a、b使得其最大公因数和最小公倍数分别为G、L。
若存在多组a、b，输出a最小的一组。
若不存在这种a、b，输出-1。

Analysis

在没有认真读题的时候，觉得这个可能要暴力枚举判断。但既然是a最小的一组，那必然是G、L本身，前提是L%G == 0。

Code

//////////////////////////////////////////////////////////////////////
//Target: UVa 11388 - GCD LCM
//@Author: Pisceskkk
//Date: 2019-2-16
//////////////////////////////////////////////////////////////////////

#include<cstdio>
#define ll long long
using namespace std;

ll a,b,l,g,d;
int t;
bool ok;

int main(){
    scanf("%d",&t);
    while(t--){
        ok = 1;
        scanf("%lld %lld",&g,&l);
        if(l%g != 0){
            printf("-1
");
            continue;
        }
        else {
            printf("%lld %lld
",g,l);
        }
    }
    return 0;
}