Problem Description

people in USSS love math very much, and there is a famous math problem .

give you two integers n ,a ,you are required to find 2 integers b ,c such that an +bn=cn .

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Input

one line contains one integer T ;(1≤T≤1000000)

next T lines contains two integers n ,a ;(0≤n≤1000 ,000 ,000,3≤a≤40000)

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Output

print two integers b ,c if b ,c exits;(1≤b,c≤1000 ,000 ,000) ;

else print two integers -1 -1 instead.

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Sample Input

1 2 3

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Sample Output

4 5

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費(fèi)馬大定理在加上勾股數(shù)的一些規(guī)律：

費(fèi)馬大定理：

a^n+b^n=c^n在n>2時(shí)是不成立的。

勾股數(shù)規(guī)律：

https://wenku.baidu.com/view/8282f1b669eae009591bec85.html

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#include <iostream>
using namespace std;

int main() ? ? ?
{
?? ?int T;
?? ?scanf("%d",&T);
?? ?int n,a,b,c;
?? ?while(T--)
?? ?{
?? ??? ?scanf("%d %d",&n,&a);
?? ??? ?if(n>2||n==0)
?? ??? ?{
?? ??? ??? ?printf("-1 -1\n");
?? ??? ??? ?continue;
?? ??? ?}
?? ??? ?else if(n==1)
?? ??? ?{
?? ??? ??? ?printf("%d %d\n",1,a+1);
?? ??? ??? ?continue;
?? ??? ?}
?? ??? ?else
?? ??? ?{
?? ??? ??? ?if(a%2==0) ? ? ? ? ?//a為偶數(shù)的時(shí)候
?? ??? ??? ?{
?? ??? ??? ??? ?int temp=(a-2)/2;
?? ??? ??? ??? ?printf("%d %d\n",temp*temp+2*temp,temp*temp+2*temp+2);?
?? ??? ??? ??? ?continue;
?? ??? ??? ?}?
?? ??? ??? ?else
?? ??? ??? ?{
?? ??? ??? ??? ?int temp=(a-1)/2;
?? ??? ??? ??? ?printf("%d %d\n",2*temp*temp+2*temp,2*temp*temp+2*temp+1);
?? ??? ??? ??? ?continue;
?? ??? ??? ?}
?? ??? ?}
?? ?}?? ??? ??? ??? ?
?? ?return 0;?? ?
}?

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