传送门

解析：

这道题由于模数一定是质数，所以我们只需要特判掉模数为2的情况，剩下的就是模奇质数二次剩余了。

关于二次剩余可以看我的博客

代码：

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define re register
#define gc getchar
#define pc putchar
#define cs const 
#define complex COMPLEX

namespace IO{
	inline int getint(){
		re int num;
		re char c;
		while(!isdigit(c=gc()));num=c^48;
		while(isdigit(c=gc()))num=(num+(num<<2)<<1)+(c^48);
		return num;
	}
	inline void outint(ll a){
		static char ch[23];
		if(a==0)pc('0');
		while(a)ch[++ch[0]]=a-a/10*10,a/=10;
		while(ch[0])pc(ch[ch[0]--]^48);
	}
}
using namespace IO;

inline ll mul(cs ll &a,cs ll &b,cs ll &mod){
	return (a*b-(ll)((long double)a/mod*b)*mod+mod)%mod;
}

struct complex{
	ll x,y;
	complex():x(1),y(0){}
	complex(ll _x,ll _y):x(_x),y(_y){}
};
ll W,mod;
complex operator*(cs complex &a,cs complex &b){
	return complex((mul(a.x,b.x,mod)+mul(mul(a.y,b.y,mod),W,mod))%mod,(mul(a.x,b.y,mod)+mul(a.y,b.x,mod))%mod);
}

inline ll quickpow(ll a,ll b){
	ll ans=1;
	for(;b;b>>=1,a=mul(a,a,mod))if(b&1)ans=mul(ans,a,mod);
	return ans;
}


inline complex quickpow(complex a,ll b){
	complex res;
	for(;b;b>>=1,a=a*a)if(b&1)res=res*a;
	return res;
}

inline ll solve(ll a){
	if(mod==2)return 1;
	if(quickpow(a,(mod-1)>>1)==mod-1)return -1;
	re ll b;
	while(true){
		b=rand()%mod;
		W=(mul(b,b,mod)-a+mod)%mod;
		if(quickpow(W,(mod-1)>>1)==mod-1)break;
	}
	return quickpow(complex(b,1),(mod+1)>>1).x;
}

int T;ll a;
signed main(){
	T=getint();
	srand(time(0));
	while(T--){
		a=getint();mod=getint();
		a%=mod;
		a=solve(a);
		if(~a){
			re ll b=mod-a;
			if(a>b)swap(a,b);
			if(a^b)outint(a),pc(' '),outint(b),pc('\n');
			else outint(a),pc('\n');
		}
		else puts("No root");
	}
	return 0;
}