Title:

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 a^n+b^n=c^n.

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)

Output

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

else print two integers -1 -1 instead.

Title:

Here are two numbers for you. N is the index, a is the base. Find b and c to satisfy a^n+b^n=c^n.
In fact, it's Fermat's theorem. When n > 2, there is no integer, and = 2 is the Pythagorean number. Here we investigate the construction of Pythagorean number,
In the original formula:

a is odd 2n+1, b = 2n^2+2n, c = b+1;

When a is even 2n greater than 4, b = n^2-1, c = n^2+1;

In this way, Pythagorean number can be constructed, but this problem is T....
So, later, by constructing the original number of shares, preprocessing, it passed.
Preprocessing Code:

void ppt(){
    ll s,t;
    for(s = 3; s <= 40000; s += 2){ 
        for(t = 1; t < s; t += 2){
            if(__gcd(s,t) == 1) {
                ll a = s*t; 
                if(a > 40000)
                    break;
                ll b = (s*s-t*t)/2; 
                ll c = (s*s+t*t)/2;
                A[a][0] = b;
                A[a][1] = c;
                if(b > 40000)
                    break; 
                A[b][0] = a;
                A[b][1] = c;
            }
        }
    }
}

Code:

#include<iostream>
using namespace std;

typedef long long ll;
int main()
{
    int t;
    ll n,m;
    while(cin >> t){
cin >> n >> m;
        if(n > 2)
            cout << -1 << " " << -1 << endl;
        else{
            if(n == 0){
                cout << -1 << " " << -1 << endl;
            }
            if(n == 1)
            {
                cout << 1 <<" " << m+1 << endl;
            }
            if(n == 2){
                if(m <= 2){
                    cout << -1 << " " << -1 << endl;
                }
                else{
                    int x = 0;
                    if(m % 2){
                        x = (m-1)/2;
                        cout << 2*x*x+2*x << " " << 2*x*x+2*x+1 << endl;
                    }
                    else{
                        x = m/2;
                        cout << x*x-1 << " " << x*x+1 << endl;
                    }

                }
            }
        }
    }
}