#include<bits/stdc++.h>

using namespace std;

typedef long long ll;

const ll mod=;

ll qp(ll a, ll b, ll c)

{

    ll ans = ;

    while (b)

    {

        if (b %  == )ans = (ans*a) % c;

        b /= ;

        a = (a*a) % c;

    }

    return ans;

}

ll p=mod;

ll w;//二次域的D值

bool ok;//是否有解

struct QuadraticField//二次域

{

    ll x, y;

    QuadraticField operator*(QuadraticField T)//二次域乘法重载

    {

        QuadraticField ans;

        ans.x = (this->x*T.x%p + this->y*T.y%p*w%p) % p;

        ans.y = (this->x*T.y%p + this->y*T.x%p) % p;

        return ans;

    }

    QuadraticField operator^(ll b)//二次域快速幂

    {

        QuadraticField ans;

        QuadraticField a = *this;

        ans.x = ;

        ans.y = ;

        while (b)

        {

            if (b & )

            {

                ans = ans*a;

                b--;

            }

            b /= ;

            a = a*a;

        }

        return ans;

    }

};

ll Legender(ll a)//求勒让德符号

{

    ll ans=qp(a, (p - ) / , p);

    if (ans +  == p)//如果ans的值为-1，%p之后会变成p-1。

        return -;

    else

        return ans;

}

ll Getw(ll n, ll a)//根据随机出来a的值确定对应w的值

{

    return ((a*a - n) % p + p) % p;//防爆处理

}

ll solve(ll n)

{

    ll a;

    if(n==)

        return ;

    if (p == )//当p为2的时候，n只会是0或1，然后0和1就是对应的解

        return n;

    if (Legender(n) == -)//无解

        ok = false;

    srand((unsigned)time(NULL));

    while ()//随机a的值直到有解

    {

        a = rand()%p;

        w = Getw(n, a);

        if (Legender(w) == -)

            break;

    }

    QuadraticField ans,res;

    res.x = a;

    res.y = ;//res的值就是a+根号w

    ans = res ^ ((p + ) / );

    return ans.x;

}

int main()

{

    ll q;

    scanf("%lld",&q);

    ll a,b,n,ans1,ans2;

    ll inv2=qp(,mod-,mod);

    while (q--)

    {

        scanf("%lld%lld",&a,&b);

        ok = true;

        n=(a*a%mod-*b%mod+mod)%mod;

        ans1 = solve(n);

        ans2 = p - ans1;//一组解的和是p

        if (!ok)

        {

            cout<<-<<' '<<-<<'\n';

        }

        else

        {

            ll x=(ans1+a)%p*inv2%p;

            ll y=(a-x+p)%p;

            printf("%lld %lld\n", min(x,y),max(x,y) );

        }

    }

}