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

give you two integers nn,aa,you are required to find 22 integers bb,cc such that anan+bn=cnbn=cn.

Input

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

next TT lines contains two integers nn,aa;(0≤n≤1000(0≤n≤1000,000000,000,3≤a≤40000)000,3≤a≤40000)

Output

print two integers bb,cc if bb,cc exits;(1≤b,c≤1000(1≤b,c≤1000,000000,000)000);

else print two integers -1 -1 instead.

Sample Input

1 2 3

Sample Output

4 5

根据费马大定理的结论，用奇偶构造法求出勾股定理中另外两个数。

AC代码：

`#include<bits/stdc++.h>using namespace std;int main(){  int t;  cin>>t;  while(t--) {    int n,a;    scanf("%d%d",&n,&a);    if(n > 2 || n==0) {      puts("-1 -1");    }    else if(n == 1) {      printf("1 %d\n",a+1);    }    else {      if(a&1) {        int tmp = (a-1)/2;        printf("%d %d\n",2*tmp*tmp + 2*tmp,2*tmp*tmp+2*tmp+1);      }      else {        int tmp = a/2 - 1;        printf("%d %d\n",tmp*tmp + 2*tmp,tmp*tmp + 2*tmp + 2);      }    }  }    return 0 ;}`